'.  II  ) 


RKEIIY 

*.      .;y  Of 
UlfOKNIA 


NEW  MATHEMATICAL   WORKS. 


An  Elemeiitaiy  Treatise  on  the  Differential  otid  Integral  Calculus. 

]{y  (J.  W.  1[i;mmin(;,  M.A.,  Ffllow  of  St.  John's  College,  C'anihiidpc, 
bvo.  bds.  9s. 

Elementaiy  Mechanics,  accompanied  by  numerous  Examples  solved 

Geometrically.  iJy  J.  15.  Tnu-ui,  M.A.,  Fellow  and  Mathematical  Lec- 
turer of  Clare  Hail.     8vo.  bdt.  10*.  6rf. 

A  Short  and  Easy  Course  of  Algebra. 

Cliicfly  designed  for  tlie  use  of  tlic  Junior  Classes  in  Schools,  -with  a 
numerous  collection  of  Original  Easy  Exercises.  By  the  Rev.  T.  Lund, 
H.D.,  late  Fellow  of  St.  .John's  College.     12mo.  bds.  3a.  6d. 

"  His  definitions  are  admirable  for  their  simplicity  and  clearness." 

Athenjeum. 

"  In  order  to  ascertain  how  far  the  Author's  performance  comes  nj)  to  fiis 
design,  ire  /tate  paid  particular  attention  to  those  places  where  the  learner 
is  most  likely  to  stumble  xtpon  acknowledged  difficulties.  ,  .  ,  In  all 
these  toe  have  much  reason  to  admire  the  happy  art  of  the  Azithor  in 
making  crooked  things  straight,  and  rough  places  sfnoofh.  The  Student 
must  be  hopelessly  obtuse  rcho  does  not,  in  folloxcing  the  guidance  of 
Mr.  Lund,  obtain  increasing  light  and  satisfactimi  in  every  step  of  his 
way;  and  such,  too,  is  the  strictly  scientific  as  well  as  simple  nature 
of  the  course  pursued,  that  he  who  makes  hitnself  master  of  it,  will 
have  laid  a  firm  foundation  for  an  extensive  and  lofty  superstructure 
of  mathematical  acquirement,", — The  Educator. 

An  Elementaiy  Treatise  on  the  Differential  Calculus. 

15y  I.  TuDiiuNTEu,  M.A.,  Fellow  of  St.  John's  College.    About  Christmas . 
This  Work  is  intended  for  the  use  of  Schools  as  well  as  for  Students  in 
the  Universities. 

Plane  Astronomy. 

Including  Explanations  of  Celestial  Phenomena  and  Descriptions  of 
Astronomical  Instruments.  By  the  Kcv.  A.  R.  Grant,  M.A.,  FelloAv 
of  Trinity  College.     8vo.  bds.  6s. 

The  Principles  of  the  Solutions  of  the  Senate -House  "  Riders" 

Exemplified  in  the  Solution  of  those  proposed  in  the  years  1848  and  1851, 
By  F.  J.  Jameson,  B.A.,  Caius  College.  In  October, 

A  Treatise  on  Dynamics. 

By  W.  P.  Wilson,  M.A.,  Fellow  of  St.  John's  College,  and  Professor 
of  Mathematics  iji  Queen's  College,  Belfast.     8vo.  bds,  9s.  Gd, 

Cambridge:   3IACMILLAN  AND   CO. 
London:    GEORGE' BELL. 


SOLUTIONS  OF  THE  CAMBRIDGE 


SENATE-HOUSE  PROBLEMS 


FOR  FOUR  YEARS  :-1848-51. 


N.  M.  FERRERS,   B.A., 

OF    GOXVILLE    AND    CAIUS    COLLEGE,    CAMBRIDGE, 
AND 

J.    STUART    JACKSON,    B.A., 

ALSO   OF   GONVILLE   AND    CAIUS    COLLEGE, 


arambrtbge:   MACMILLAN   &  Co.; 

lEontJon:   GEOKGE  BELL;     ©ubiin:  HODGES   &   SMITH; 

^(nbnrg^ :    EDMOXSTON  &  DOUGLAS  ; 

(Glasgoto :    J.  MACLEHOSE. 

1851. 


LOAN  STACK 


CAMBRIDGE: 

PHINTKD    I!Y    MF.TCAI.FE    AND    PALMl.I'. 


PREFACE. 

It  will,  we  believe,  be  universally  admitted  that  there  is 
no  easier  means  of  becoming  acquainted  with  any  branch  of 
Mathematics,  than  the  study  of  Examples  illustrative  of  its 
principles.  It  is  also  indispensably  necessary  that  the  ingenuity 
of  the  Student  be  thoroughly  exercised  in  attempting  to  dis- 
cover for  himself  the  solution  of  any  problem  which  may  be 
put  before  him :  it  is  by  no  means  our  object,  in  publishing 
this  book,  to  save  him  the  trouble  of  doing  so.  But  we  believe 
that  if,  after  having  done  his  best  to  master  a  problem  for 
himself,  he  is  still  unsuccessful,  he  will  then  derive  great  benefit 
from  referring  to  the  solution  obtained  by  another  person.  It 
is  for  this  pui'pose  that  we  hope  the  present  collection  will  be 
foimd  of  service. 

Of  the  intrinsic  value  of  these  Problems  we  could  have  no 
doubt,  even  if  we  knew  less  of  them,  coming  as  they  do  from 
such  high  authorities.  We  have  as  little  doubt  that  they  are 
on  all  accounts  the  best  problems  that  we  could  have  chosen 
for  solution  for  the  benefit  of  the  Student,  for  their  general 
value,  their  variety,  and  because  they  shew  what  Senate- 
house  Problems  are,  and  arc  likely  to  be  in  coming  years. 


418 


IV  PREFACE. 

Part  1.  contains  the  solutions  of  those  proposed  in  tlic  first 
three  days  of  examination :  they  are  of  a  simpler  kind  than 
those  proposed  in  the  remaining  five  days,  the  solutions  of  which 
form  Part  II.  llie  solutions  of  many  problems  have  been 
kindly  timiished  by  the  Moderators  by  whom  they  were  pro- 
posed: we  take  this  opportunity  of  returning  om*  acknowledg- 
ments to  them  and  others  of  our  friends  who  have  assisted 
us  in  the  progress  of  the  work.  We  shall  also  feel  much 
obliged  to  any  of  our  readers  who  will  send  us  either  correc- 
tions of  our  solutions  or  improvements  upon  them. 

Some  difficulty,  it  will  easily  be  understood,  has  been  found 
in  bringing  all  the  problems  to  appear  m  their  right  places: 
any  problems,  however,  which  have  been  omitted  in  the  body 
of  the  work,  will  be  found  in  the  Appendix,  The  problems 
on  pp.  226,  241,  should  have  been  placed  among  the  Trigo- 
nometry of  1850,  and  the  Geometry  of  Three  Dimensions  of 
1851,  respectively. 


C  O  IS  T  E  N  T  S. 


PART   I. 


Euclid    . 
Algebra 
Trigonometry 
Conic  Sections 
Statics    . 
Dynamics 
Newton 
Hydrostatics 
Optics    . 
Astronomv    • 


Page 
1 

9 
19 
25 
32 
37 
48 
53 
59 
64 


PART   II. 


Euclid    . 
Algebra 

Plane  Trigonometry 
Spherical  Trigonometry 
Theory  of  Equations 
Geometry  of  Two  Dimensions 
Differential  Calculus 
Integral  Calculus 


69 
74 
98 
107 
112 
110 
171 
183 


i  CONTUNTS. 

Page 

Geometry  oH  Three  Dimensions           .                 .                 .  1*«^3 

Ditferential  Equations                   ....  222 

Definite  Integrals                  ....  233 

Calculus  of  Finite  Differences     ....  240 

Statics                     .....  245 

Dynamics  of  a  Particle                ....  2*37 

Rigid  Dynamics    .                 .                .                 .                 •  283 

Hydrostatics                  .....  323 

Hydrodynamics     .....  333 

Geometrical  Optics       .....  339 

Astronomy             .                 .                 •                 •                 •  3J5 

Disturbed  Motion         .....  362 

Attractions            .....  36j 

Physical  Ojjtics            .....  375 

Calculus  of  Variations          ....  380 

Appendix     ......  384 


PART  I. 


SOLUTIONS  OF  THE  SENATE-HOUSE  PROBLEMS. 


EUCLID. 


ERRATA. 

I'AGK       LINE  /.  ;  J      ,.' 

87,  5  from  bottom,  for  p  read  n . 

95,  10,  for  (i,,.,n"p  ^''^"'■^  "7'---i  +  "^• 

—  17,  for  ...  t  <v.,)-i  Vi^  '■'"'^^  •■•■^  "p  0   Vi- 

176,  12,  /or  /t'  -  2rt/  rcat^  A'  -  2a A. 

288,  8,"/o?-  -R'  J-ertc/  By. 

215,  1,  supply  reference  to  figure  (S8). 

23l',  !,>'•  (tig- 08)  7-m(/ (tig.  89). 

Figure  4f5,  /b;-  j>  2/  s  »e'^t^  ~ i^  y- 


Again,  because  EF  is  a  diameter  of  the  circle,  therefore  the 
angle  FCE  is  a  right  angle.  But  CE  bisects  the  inght  angle 
ACB,  therefore  ACE  is  half  a  right  angle,  therefore  also  FCA 
is  half  a  right  angle ;  that  is,  FC  bisects  the  supplement  of  the 
right  angle  ACB. 

2.  A,  B,  C,  (fig.  2)  are  three  given  points  in  the  circum- 
ference of  a  circle  ;  find  a  point  P,  such  that  if  AP,  BP,  CP 
meet  the  circumference  in  D,  E,  F,  the  arcs  DE,  EF,  may  be 
equal  to  given  arcs. 

Join  AB,  and  on  it  describe  a  segment  of  a  circle,  containing 
an  angle  equal  to  the  sum  of  those  subtended  at  the  circiim- 


SOLUTIONS  OF  THE  SENATE-HOUSE  PROBLEMS. 


EUCLTD. 

1848. 

1.  If  the  hypotlienuse  AB  (fig.  1)  of  a  right-angled  triangle 
ABC  be  bisected  in  D,  and  EDF  drawn  pei-pendicular  to  AB, 
and  DE,  DF  cut  off  each  equal  to  DA,  and  CE,  CF  joined ; 
prove  that  the  last  two  lines  will  bisect  the  angle  at  C  and  its 
supplement  respectively. 

Join  CD,  then  shall  CD  be  equal  to  half  the  hvpothenuse 
AB,  that  is,  to  DE  or  DF ;  therefore  a  circle  described  from 
centre  D,  with  radius  DC,  will  pass  through  B,  E,  A,  F.  Let 
this  circle  be  described,  then  the  angle  ECB,  at  the  circum- 
ference, is  equal  to  half  the  angle  EDB  at  the  centre,  that  is 
to  half  a  right  angle,  and  therefore  to  half  the  angle  ACB  ; 
that  is,  CE  bisects  the  angle  ACB. 

Again,  because  EF  is  a  diameter  of  the  circle,  therefore  the 
angle  FCE  is  a  right  angle.  But  CE  bisects  the  right  angle 
ACB,  therefore  ACE  is  half  a  right  angle,  therefore  also  FCA 
is  half  a  right  angle  5  that  is,  FC  bisects  the  supplement  of  the 
right  angle  ACB. 

2.  A,  B,  C,  (fig.  2)  are  three  given  points  in  the  circum- 
ference of  a  circle  ;  find  a  point  P,  such  that  if  AP,  BP,  CP 
meet  the  circmnference  in  D,  E,  F,  the  arcs  DE,  EF,  may  be 
equal  to  given  arcs. 

Join  AB,  and  on  it  describe  a  segment  of  a  circle,  containing 
an  angle  equal  to  the  sum  of  those  subtended  at  the  circum- 

H 


2  SOLUTION    OF   SENATE-HOUSE    PHOBLEMS.  [1849. 

forence  by  the  arcs  AB  and  DE.  Also  join  BC,  and  on  it 
describe  a  segment  of  a  circle  containing  an  angle  equal  to  the 
sum  of  those  subtended  at  the  circumference  by  the  arcs  BC 
and  EF.  These  segments  shall  intersect  in  the  required  point  P. 
For  join  AP,  BP,  CP,  and  produce  them  to  meet  the  cir- 
cumference in  D,  E,  F,  respectively.  Join  AE,  then  the  angle 
EAD  is  equal  to  the  difference  of  the  angles  APB,  AEP, 
that  is,  to  the  angle  required  to  be  subtended  by  the  arc  DE. 
Therefore  DE  is  equal  to  the  arc  required.  Similarly  it  may  be 
shewn  that  EF  is  equal  to  the  arc  required,  and  therefore  P  is 
the  required  point. 

1849. 

1.  Through  a  point  C  (fig.  3)  in  the  eirciunference  of  a  circle, 
two  straight  lines  ACB,  DCE,  are  drawn,  cutting  the  circle  in 
B  and  E  ;  prove  that  the  straight  line  which  bisects  the  angles 
ACE,  DCB,  meets  the  circle  in  a  point  equidistant  from  B 
and  E. 

Let  CP  be  the  line  bisecting  the  angles  ACE,  BCD ;  P  the 
point  in  which  it  meets  the  circle.  Join  PB,  PE,  BE ;  then 
because  the  angles  PBE,  PCE  are  in  the  same  segment,  there- 
fore they  are  equal  to  one  another. 

Again,  because  the  angles  BCP,  BEP  are  opposite  angles 
of  a  quadrilateral  inscribed  in  a  circle,  therefore  they  are 
together  equal  to  two  right  angles,  that  is  to  ACP  and  BCP. 
Therefore,  taking  away  the  common  angle  BCP,  ACP  is  equal 
to  BEP.  But  ACP  is  equal  to  ECP  by  constniction,  therefore 
from  above  ACP  is  equal  to  EBP :  and  it  has  been  shewn  to  be 
equal  to  BEP,  therefore  the  angles  EBP,  BEP  are  equal  to 
one  another,  therefore  PE  is  equal  to  PB.  That  is,  the  point  P 
in  which  the  bisecting  line  CP  meets  the  circle  is  equidistant 
from  B  and  E. 

2.  Two  circles  intersect  in  A  and  B  (fig.  4).  At  A,  the 
tangents  AC,  AD  are  drawn  to  each  circle  and  tenninated  by 
the  circmnference  of  the  other.  If  BC,  BD  be  joined,  shew 
that  AB,  or  AB  produced  if  necessary,  bisects  the  angle  CBD. 


1849.]  EUCLID.  3 

Produce  CA,  DA,  to  E,  F.  Then  the  angle  CAF  is  equal 
to  the  angle  DAE :  but  the  angle  CAF  is  equal  to  the  angle 
ABC  in  the  alternate  segment,  also  the  angle  DAE  is  equal  to 
the  angle  ABD  in  the  alternate  segment.  Therefore  the  angles 
ABC,  ABD  are  equal  to  one  another,  and  AB,  produced  if 
necessary,  bisects  the  angle  CBD. 

3.  Draw  a  line  to  touch  one  given  circle,  so  that  the  part 
of  it  contained  by  another  given  circle  may  be  equal  to  a  giveii 
straight  line,  not  greater  than  the  diameter  of  this  latter  circle. 

Let  ABC,  DEF  (fig.  5)  be  two  given  circles ;  it  is  required 
to  draw  a  straight  line  touching  the  circle  ABC,  so  that  the 
part  of  it  contained  by  DEF  may  be  equal  to  a  given  straight 
line,  not  greater  than  the  diameter  of  DEF. 

In  the  circle  DEF  place  the  straight  line  DE,  equal  to  the 
given  straight  line.  Find  G  the  centre  of  this  circle,  and  with 
G  as  centre  describe  a  circle  touching  DE.  Draw  AFH  a 
common  tangent  to  this  latter  circle  and  ABC,  cutting  DEF 
in  F,  H,  this  shall  be  the  line  required. 

For  since  FH,  DE  each  touch  a  circle  whose  centre  is  G, 
therefore  they  are  equidistant  from  G,  the  centre  of  the  circle 
DEF.     Therefore  FH  is  equal  to  DE. 

Hence  AFH  is  drawn  touching  the  cii-cle  ABC,  and  the 
part  of  it  contained  by  DEF  is  equal  to  the  given  straight  line. 

4.  A  quadrilateral  figure  possesses  the  following  property : 
any  point  being  taken,  and  four  triangles  fonned  by  joining  this 
point  with  the  angular  points  of  the  figure,  the  centres  of 
gravity  of  these  triangles  lie  in  the  circumference  of  a  circle ; 
prove  that  the  diagonals  of  this  quadrilateral  arc  at  right  angles 
to  each  other. 

Let  ABCD  (fig.  6)  be  the  quadrilateral,  P  any  point  within 
it :  E,  F,  H,  K,  the  middle  points  of  the  sides.  Join  PE,  PF, 
PH,  PK.  And  in  them  take  PG„  PG,,  PG„  PG,,  respectively 
equal  to  two-thirds  of  PE,  PF,  PH,  PK.  Gfi.jGjOc^  shall  be 
the  centres  of  gravity  of  the  triangles  PAB,  PBC,  PCD,  PDA. 
Join  EF,  FH,  HK,  KE,  these  lines  are  evidently  parallel  to 

b2 


4  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

^^.^\i  C^^Gs,  G3G,,  G,G, ;  and  therefore  E,  F,  H,  K,  lie  in  the 
cireumference  of  the  same  circle.  But  EF,  HK  are  each 
parallel  to  the  diagonal  AC,  therefore  also  to  each  other. 
Similarly  FH,  EK  arc  parallel  to  each  other,  therefore  EFHK 
is  a  parallelogram.  And  since  it  is  inscribable  in  a  circle,  each 
of  its  angles  is  a  right  angle.  Therefore  also  the  diagonals 
AC,  BD,  which  are  respectively  parallel  to  the  sides  of  the 
parallelogram,  are  at  right  angles  to  each  other. 

1850. 

1.  If  ABCD  (fig.  7)  be  a  parallelogram,  and  P,  Q  two 
points  in  a  line  parallel  to  AB,  and  if  PA,  QB  meet  in  E, 
and  PD,  QC  in  S,  prove  that  RS  is  parallel  to  AD. 

Because  CD  is  parallel  to  QP,  therefore  SD  is  to  SP  as  CD 
to  PQ.  And  because  AB  is  parallel  to  PQ,  therefore  RA  is  to 
RP  as  AB  to  PQ.  But  AB  is  equal  to  CD,  therefore  RA  is  to 
RP  as  SD  to  SP,  therefore  RP  is  to  AP  as  SP  to  DP,  there- 
fore RS  is  parallel  to  AD. 

2.  Two  sides  of  a  triangle,  whose  perimeter  is  constant,  are 
given  in  position ;  shew  that  the  third  side  always  touches  a 
certain  circle. 

Let  ABC  (fig.  8)  represent  the  triangle ;  AB,  AC  being  the 
sides  given  in  position.  Describe  a  circle  DEF  touching  BC 
and  AB,  AC  produced.  The  side  BC  shall  always  touch  the 
circle  DEF. 

For  since  BD,  BF  both  touch  the  same  circle,  therefore  BD 
is  equal  to  BF.     Hence  AD  is  equal  to  AB,  BF  together. 

Similarly  AE  is  equal  to  AC,  CF  together. 

Therefore  AD,  AE  together  are  equal  to  AB,  AC,  BC 
together ;  that  is,  to  the  perimeter  of  the  triangle  ABC,  which 
is  constant.  But  since  AD,  AE  touch  the  same  circle,  therefore 
AD  is  equal  to  AE,  and  their  sum  has  been  shewn  to  be  con- 
stant; therefore  AD,  AE  are  each  constant,  that  is,  the  circle 
touching  BC,  and  AB,  AC  produced,  touches  AB,  AC  in  fixed 
points ;  that  is,  it  is  a  fixed  circle.  Therefore  BC  always 
touches  a  fixed  circle. 


1851.]  EUCLID.  5 

1851. 

1.  In  AB,  the  diameter  of  a  circle,  take  two  points  C,  D, 
equally  distant  from  the  centre,  and  from  any  point  E  in  the 
circmnference  draw  EC,  ED ;    shew  that 

EC'  +  ED''  =  AC  +  AD^ 

Take  O  (fig.  9)  the  centre  of  the  circle,  and  join  EO,  and 
di'aw  EF  pei'pendicular  to  AB. 

Then  because  CD  is  bisected  in  O,  and  produced  to  A, 

.-.  AC'  +  AD'  =  2  (AC  +  OC)   [Em.  ii.  10). 

Again,  because  EC  is  opposite  to  an  acute  angle  O  of  a  triangle 
ECO,  therefore 

EC  +  20C.0F  =  EO'  +  OC  [Euc.  ii.  13). 

And  because  ED  is  opposite  to  the  obtuse  angle  O  of  a  triangle 
EOD,  therefore 

ED''  =  EO'  +  OD'  +  20D.0F   [Euc.  ii.  12), 

=  EO'  +  OC  +  20C.0F ; 

.-.  EC^  +  ED'  =  2  (EC  +  OC), 
=  2(A0'  +  0C), 
=  AC  4-  AD'  from  above. 

2.  If  through  the  fixed  points  P,  Q,  (fig.  10)  parallel  lines  be 
drawn  meetmg  two  fixed  parallel  lines  in  the  points  M,  N ; 
then  the  line  through  the  points  M,  N,  passes  through  a  fixed 
point. 

Join  PQ,  and  let  it  meet  MN  in  O,  and  the  given  pair  of 
parallels  in  Ii,  S,  O  shall  be  a  fixed  point. 

For  since  QN  is  parallel  to  PM,  therefore  QO  is  to  PQ  as 
ON  to  NM.  And  since  NR  is  parallel  to  MS,  therefore  OR  is 
to  RS  as  ON  to  NM.  Therefore  OR  is  to  RS  as  OQ  to  QP, 
or  OR  is  to  OQ  as  RS  to  QP ;  that  is,  QR  is  divided  in  a  con- 
stant ratio  in  O,  and  therefore  O  is  a  fixed  point. 

3.  In  a  given  circle  it  is  required  to  inscribe  a  triangle, 
similar  and  similarly  situated  to  a  given  triangle. 


6  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Let  ABC  (rig.  11)  be  a  given  triangle,  DEF  a  given  circle; 
it  is  required  to  inscribe  in  DEF  a  triangle,  similar  and  similarly 
situated  to  ABC. 

At  the  point  A  in  the  straight  line  AB,  make  the  angle 
BAG  equal  to  the  angle  ACB.  Find  H,  the  centre  of  the 
circle  DEF,  and  draw  UK  parallel  to  AG,  HD  pei*pcndicular 
to  HK.  Through  D  draw  DE,  DF  parallel  respectively  to 
AB,  AC,  DEF  shall  be  the  triangle  required. 

For  di'aw  DL  parallel  to  HK  or  AG,  and  therefore  touching 
the  circle  at  D.  Then  the  angle  LDE  is  equal  to  the  angle 
GAB.  But  GAB  is  by  constniction  equal  to  ACB,  and  LDE 
is  equal  to  DFE  in  the  alternate  segment.  Therefore  the  angle 
DFE  is  equal  to  ACB.  Similarly  the  angle  DEF  is  equal  to 
the  angle  ABC.  Therefore  the  remaining  angle  EDF  is  equal 
to  BAC,  so  that  the  triangles  ABC,  DEF  are  similar.  And 
since  the  sides  DE,  DF  are  parallel  respectively  to  AB,  AC, 
therefore  EF  is  parallel  to  BC,  and  they  are  similarly  situated. 

4.  Describe  a  circle  which  shall  pass  through  two  points, 
and  cut  off  from  a  given  straight  line  a  chord  of  given 
length. 

Let  A,  B  (fig.  12)  be  two  given  points,  CX  a  given  line, 
it  is  required  to  describe  a  circle  passing  tlu'ough  A,  B,  and 
cutting  off  from  CX  a  chord  of  given  length. 

Join  BA,  and  produce  it  to  meet  CX  in  C.  Bisect  AB  in  E, 
and  draw  EF  perpendicular  to  AB.  With  A  as  centre,  and 
radius  equal  to  half  the  required  chord,  describe  a  circle  FGH, 
cutting  EF  in  F.  With  F  as  centre,  and  FA  as  radius,  describe 
a  circle  BAG.  Join  CF,  and  let  it  cut  the  circle  BAG  in  K,  L. 
From  CX  cut  off  CM,  equal  to  CK.  The  circle  described 
through  A,  B,  ]\I  shall  be  the  circle  required ;  that  is,  if  N  be 
the  second  point  in  which  it  meets  CX,  MN  shall  be  equal  to 
the  required  chord. 

For  the  rectangle  CM.CN  is  equal  to  CA.CB  by  the  pro- 
perty of  the  circle  ABM.  And  the  rectangle  CA.CB  Is  equal 
to  CK.CL  by  the  property  of  the  circle  KAB.  Therefore  the 
rectangle  CK.CL  is  equal  to  CM.CN. 


1851.]  EUCLID.  7 

But  CK  is  equal  to  CM,  therefore  CL  is  equal  to  CN,  and 
therefore  KL  is  equal  to  MN,  and  KL  is  equal  to  the  required 
chord,  therefore  MN  is  so,  and  the  circle  ABNM  is  the  circle 
required. 

5.  Give  a  constniction*  for  finding  the  common  tangents  of 
two  circles,  and  shew  that  if  through  the  intersection  O  of  two 
of  the  common  tangents  which  meet  in  the  line  joining  the 
centres  of  the  two  circles,  there  be  drawn  a  transversal  meeting 
the  circles  in  A,  A',  and  B,  B',  respectively,  then  (the  points 
denoted  by  B,  B'  being  properly  chosen)  OA.OB'  =  OA'.OB  is 
independent  of  the  position  of  the  transversal. 

Let  ABC,  DEF  (fig.  13)  be  two  circles,  whereof  DEF  is  the 
greater;  find  G,  H,  their  centres,  and  with  H  as  centre,  and 
radius  equal  to  the  differencef  of  the  radii  of  the  given  circles, 
describe  a  circle.  Through  G  draw  GK,  a  tangent  to  this 
circle ;  di-aw  GA  perpendicular  to  GK,  and  AD  perpendicular 
to  GA,  meeting  DEF  in  D :  AD  shall  be  a  common  tangent. 

Let  C,  D  (fig.  14)  be  the  points  of  contact  of  one  of  the 
coimnon  tangents.     Then  we  easily  see  that 
OA:OA'  ::  OB:  OB', 
.-.     OA.OB'  =  OA'.OB  ; 
and  also       OA.OB'  :  OA.OA  ::  OB.OB'  :  OA'.OB, 
or  OA.OB'  :  OC^:  OD^  :  OA'.OB, 

::0D^  OA.OB', 
.-.    OA.OB'  =  OA'.OB  =  OC.OD, 
which  is  independent  of  the  position  of  the  transversal. 

6.  Shew  that  a  triangle  made  to  revolve  in  the  same  direc- 
tion about  its  three  angular  points  in  a  proper  order  through 
angles  double  of  the  angles  of  the  triangle  at  the  same  angular 
points,  will  return  to  its  original  position. 

*  A  demonstration  of  this  construction  is  not  required. 

t  We  might  also  take  a  radius  equal  to  the  sum  of  the  radii  of  the  given 
circles,  in  which  case  the  common  tangent  would  touch  the  two  circles  on 
opposite  sides  of  the  line  joining  their  centres. 


8  SOLUTIONS   OF   SENATE-HOUSE   PKOBLEMS.  [1851. 

Let  ABC  (fig.  15)  denote  the  triangle  in  its  first  position. 
At  the  point  A,  in  the  straight  line  AB,  make  the  angle  BAG' 
equal  to  the  angle  BAG,  and  make  AG'  equal  to  AG.  Again, 
make  the  angle  GAB'  equal  to  the  angle  BAG',  and  AB'  equal 
to  AB.  Join  B'G' ;  then  in  the  triangles  GAB,  G'AB',  the  sides 
GA,  AB  are  respectively  equal  to  G'A,  AB',  and  the  angle 
GAB  is  equal  to  the  angle  G'AB' ;  therefore  the  triangles  are 
equal  in  all  respects.  And  the  angle  B'AB  is  double  of  the 
angle  GAB,  therefore  G'AB'  is  the  position  of  the  triangle  after 
revolving  round  A  through  an  angle  equal  to  2. GAB. 

Again,  join  BG' ;  make  the  angle  BG'A'  equal  to  the  angle 
B'G'A  or  BGA,  and  G'A'  equal  to  G'A.  Join  A'B ;  then  in  the 
triangles  AG'B',  A'G'B,  the  sides  AG',  G'B'  are  respectively 
equal  to  A'G',  G'B,  and  the  angle  AG'B'  is  equal  to  A'G'B. 
Therefore  the  triangle  A'BG'  is  equal  in  all  respects  to  AB'G', 
and  therefore  to  ABG.  And  the  angles  BG'A',  AG'B  are 
together  double  of  B'G'A  or  BGA,  therefore  A'BG'  is  the  posi- 
tion of  the  triangle  after  revolving  round  the  angles  A,  G. 

Again,  since  the  angles  G'BA',  G'BA,  GBA,  are  all  equal, 
GBG'  is  double  of  A'BG'  or  ABG.  Therefore  the  triangle,  after 
revolving  romid  its  three  angular  points  in  succession,  through 
angles  double  of  the  angles  of  the  triangle  at  those  points, 
returns  to  its  initial  position  ABG. 


(     9     ) 


ALGEBRA. 

1848. 

1.  A  ship  sails  with  a  supply  of  biscuit  for  60  days,  at  a 
daily  allowance  of  1  lb.  a-head :  after  being  at  sea  20  days  she 
encounters  a  storm,  in  which  5  men  are  washed  overboard,  and 
damage  sustained  that  will  cause  a  delay  of  24  days,  and  it  is 
found  that  each  man's  allowance  must  be  reduced  to  f  lb.  Find 
the  original  number  of  the  crew. 

Let  X  =  the  original  number  of  the  crew. 
Then  60a;  =  number  of  lbs.  of  biscuit  with  which  they  started. 

4:0a;  = remaining  after  being  at  sea 

20  days. 
And  the  remainder  of  the  crew  =  a;  —  5,  who  have  to  remain  at 
sea  for  64  days,  under  a  daily  allowance  of  f  lb  per  man. 

.-.    f  64  (a;  -  5)  =  40a;, 
.-.    64a;  -  320  =  56a;, 
.-.    8a;  =  320, 
and  X  —  40, 
the  original  number  of  the  crew. 


2.    If  a,  i,  and  x  be  positive,  and  a  >  b,  prove  that 


X  +  a  X  -\-  b 


>  < 


(a;"  +  a^)i  {x'  +  h' 

,,,    ,                      x  -\-  a  X  +  b 

W  e  have  t-^. ^.  >  < 


yjTi ,  according  as  a-  >  <  [ab) 


[x'  +  ay  (a;*  +  b'J  ' 

x^  +  2aa;  +  a"  x^  +  2bx  +  h 

as  = :, —  >  < 


X*  +  d'  x^  +  W       ' 

2aa;  ibx 

x'  +  «'  x'  +  /''  ' 


10  SOLUTIONS  OF   SENATE-HOUSE   PROBLEMS.  [1849. 

a                   h  .  .         .. 

as  -5 r.  >  <  -5 Ti ,  Since  x  is  positive, 

as   [a  —  h)  x''  >  <  d'h  —  a//, 
as   (a  —  h)  x^  >  <  ah  [a  —  Z>), 

as  x^  >  <  ai,  since  a  is  greater  than  b, 
or  as  X  >  <  («&)*. 

3.   If  a,  ^,  c,  be  in  haiinonic  progression,  and  ti  be  a  positive 
integer,  shew  that  a"  +  c"  >  25". 

Suppose  a,  5,  c,  all  positive.*     Then  we  have 

-  ,  T  1  -  •)  in  aritlinietic  progression, 
a      b     c 

2ac 

but   a  +  c  >  2\/«c, 

2ac      ^  , 
.*.  -^—  >  2v«c, 

0 

or    V^c  >  5, 

but   a"  +  c"  >  2a*"c*", 
a  fortiori   a'  +  c"  >  2V. 

1849. 

1.   Reduce  to  its  simplest  fonn  the  expression 

(1  -  gQ  (1  -  b'')  (1  -  <f)  -  (g  +  be)  [b  +  cff)  (c  +  a^) 
1  -  a'  -  ^*'  -  c'  -  2abc 

We  have  (l-a'')  (1-Z*^)  [l-c') 

=  1  _  a^  _  ^,'^  _  c^  +  J'-^c^  +  c'-'a'-'  +  o;'//  -  d'bV 
[a  +  be)  [b  +  CO)  (c  +  (ib) 

=  aba  +  ¥c'  +  6'ci'  +  c^W  +  [c^  +  5^  +  e')  abc  +  d'V'c^ ; 


*  If  one  of  these  three  quantities  were  negative,  the  proposition  enunciated 
Avould  not  be  necessarily  true. 


1850.]  ALGEBRA.  11 

...  (1  _  a')  (1  -  h'')  (1  -  6')  -{a  +  he)  [h  +  m)  (c  +  ab) 

=  l-d'-  ¥  -  c'  -  abc  -  {a'  +  1/  +  6')  abc  -  2dVc' 
=  [\-ci'-  h'  -  c'  -  2abc)  (1  +  abc), 
(1  _  d^)  (1  -  i;^)  (1  _  c')  -{a  +  be)  [b  +  m)  (c  +  ah) 
'  '  i-ce-h''-  &  -  2abc  ~  ^  +  ''^''• 

2.  Find  a  whole  number  which  is  greater  than  three  times 
tlie  integral  part  of  its  square  root  by  miity.  Shew  that  there 
are  two  solutions  of  the  problem,  and  no  more. 

Let  X  be  the  integral  part  of  the  square  root, 
x^  +  y  the  whole  number. 
Then,  by  the  conditions  of  the  problem, 
a;^  +  y  =  3a;  +  1, 
.•.  a;''  —  3a;  =  1  —  ?/, 
.-.  x'  _  3aj  +  f  =  1,3  _  y^ 

Now  y  is  essentially  positive,  and  in  order  that  x  may  be 
real  it  is  necessary  that  y  be  less  than  5'.  Also,  in  order  that  x 
may  be  an  integer,  i}^  —  y)^  must  be  of  the  form  |-(2?/i+l), 
m  being  some  integer,  such  that  [2m  +  l)'"*  is  less  than  13. 
Therefore  the  only  admissible  values  of  m  are  0  and  1.     Hence 

a;  =  i  +  ^ori  +  i, 
=  2  or  3. 
Therefore  3a;  +  1  =  7  or  10, 

and  7  and  10  are  the  only  solutions  of  the  problem. 

1850. 

1.  A  number  of  persons  were  engaged  to  do  a  piece  of  work 
which  would  have  occupied  them  m  hours  if  they  had  com- 
menced at  the  same  time;  but  instead  of  doing  so,  they  com- 
menced at  equal  intervals,  and  then  continued  to  work  till  the 
whole  was  finished,  the  payment  being  proportional  to  the  work 
done  by  each ;  the  first  comer  received  r  times  as  much  as  the 
last :  find  the  time  occupied. 


12  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

Let  X  be  the  number  of  persons  employed,  y  the  nmnber  of 

hours  the  last  worked,  z  the  interval,  in  hours,  between  the  first 

and  second  person  commencing  work,  which  is  also  that  between 

the  second  and  third,  and  so  on.     Then,  by  tlie  conditions  of  the 

problem, 

y  +  [x-l)  z  =  ry (1). 

Also,  if  W  be  the  work  done  in  an  hour, 

[?/+  {y-\-z)  +  {y+^z)  +  ...  +  {?/+  [x—\)  zW  W=  whole  work  done, 
•*•  y  +  (i/+^)  +  (y+2s)  4-  ...  +  {y  +  (a;-l)  s}  =  mx, 
or    [2y  +  (a?  —  1)  z]  \x  =  mx^ 
.'.    2y  -\-  {x  —  l)z  =  2m, 
and,  by  (1),   y  +  {x-l)  z  =  ry, 

.'.   y  =  2m  —  ry, 

Therefore  whole  time  occupied  =  time   from  the  first  person 
begiiming  to  work  till  the  completion  of  the  work 

=  y  +  {x-l)z, 

=  ry  by  (1), 

2mr    ,  ,      .  , 

= hours  by   2). 

r  +  1  -^  ^  ^ 

2.  Shew  that  the  product  of  the  terms  of  an  arithmetical 
progression  is  greater  than  [alf" ;  and  that  the  sum  of  the  terms 
of  a  geometrical  progression  is  less  than  (a+?)^«;  where  in 
both  cases  a,  /,  and  n  denote  the  first  and  last  terms,  and  the 
number  of  terms  respectively. 

(a)  Let  h  be  the  common  difference  in  the  arithmetical 
progression,  P  the  product  of  its  terms,  then 

P=a{a  +  b)  {a  +  2h)  ...  {a  +  {n-l)b}, 

and    I  =  a  +  [n  —  1)  b. 

Now  (a  +  mb)  [a -f-  {n-  )ti—l)  b}  =  a^  +  (n  — 1)  ab  +  (n—m  —  1)  mh^. 


1851.]  ALGEBRA.  13 

This  will  be  least  when  m  =  0,   negative  values  of  m  being 
excluded;  but  in  that  ease 

{a  +  mh)  [a -\-  (n  —  m  —  \)  b]  =  al- 
therefore  the  product  of  any  two  tenns  equidistant  from  the 
mean  is  not  less  than  al. 

The  property  enunciated  follows  at  once  from  this  when  n  is 
even.     If  7i  is  odd,  the  middle  term  is 

71  —  1  - 

and   (a  +  ^^-^  h\   =  «'  +  {n -  1)  ah  +  \^-^  o\  , 

>  d^  +  [n—  1)  «J, 

>  al ; 
.'.  a  +  ^-^  h  >  {aI)K 

Hence,  in  all  cases,  P  >  [al)-". 

(/3)  Let  r  be  the  common  ratio  in  the  geometrical  pro- 
gression, S  the  sum  of  its  tenns ;  then 

b  =  a  — —  , 
r  —  1  ' 

and    I  =  ar"'^^ 

.'.    a  +  l  =  a{l+  r"-'). 

Now  ar"^  +  ar"-"-'  -  (a  +  7)  =  «  ('•'"  -  1)  -  «  (^''"'  -  »•""""'), 

=  «  (r    -  1)  (1  -  r        ), 

which  is  negative  for  all  positive  values  of  m. 

Hence  the  smn  of  the  first  and  last  tenns  is  greater  than  the 
sum  of  any  other  two  tenns  equidistant  from  the  mean. 

The  property  enunciated  follows  at  once  from  this  when  n  is 
even.     If  n  be  odd,  the  middle  tenn  is 

and   ri^"-'>  <  ^^f^, 
,,„_!)      a  +  I 


2 
Hence,  in  all  cases,  S  <  (a  +  /)  ^n. 


14  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

1851. 

3.r  —  2 

1.    If  , -xV 7:^r} ttn  be  expanded  in  a  series  ascend- 

[x—  1)  [x  —  2)  [x  —  6)  ^ 

lug  by  powers  of  a:,  find  the  coefficient  of  x\ 

This  is  best  effected  by  resolving  the  given  expression  mto 
its  partial  fractions,  and  expanding  each  fraction  separately. 
For  this  piu'}30se,  assume 

3a; -2  A  B  G 

{x-l)  {x-2)[x-^)~  x-l'^  x-2'^  x-^"* 
A,  Bj  C  being  independent  of  x. 

ThenSx-2=A{x-2){x-3)  +  B{x-3){x-l)  +  C{x-l){x-2) 
Hence,  putting  a;  =1,  identicalhj. 

3.1  -  2  =  ^(1-2)  (1-3), 

or    1  =  2A (1). 

Similarly,  putting  a;  =  2, 

3.2  -  2  =  5(2-3)  (2-  1), 

or   4  =  -  B (2), 

and  putting  a^  =  3, 

3.3  -  2  =  0(3-1)  (3-2), 

or    7  =2C (3); 

Sx-2  _         1 4_  7 

•''    (a;-l)(a;-2)(a;-3)  ~  2  (a;  -  1)       a;  -  2  "^  2  (a;  -  3) ' 


11  2  7        1 

2"^  u     J.  —  J 


2   1-a;l-^a;      6  1-ia;' 

=  -i(l+a;+  ...  +a;"+...) 

+  2{l+^x-}-  ...  +  {^xY  + 
-l{l  +  ^x  +  ...  +  i^xr +  ...], 

in  which  the  coefficient  of  a;" 

_  _  1        2  _  7    1 
~~2'^2^~6   3""' 
_  J 1.  _  7     1 

-  2«-i       2       2   3^" 

the  required  coefficient. 


1851.]  ALGEBRA.  15 

2.  Find  the  siun  of  the  different  numbers  which  can  be 
formed  with  m  digits  a,  n  digits  /3,  &c.,  the  entire  series  of 
w  +  »  +  &c.  digits  being  employed  in  the  formation  of  each 
number. 

The  total  number  of  numbers  which  can  thus  be  fonned,  is 

equal  to  the  number  of  permutations  of  m  -^-n  -\- things, 

whereof  wi  are  of  one  kind,  n  of  another...,  taken  all  together, 

_    1.2...(m  +  /i  +  ...) 

~  (1.2. ..w)  (1.2...??)...  ' 

Now,  the  number  of  times  a  will  be  found  in  any  assigned 

place  :  the  nmnber  of  times  j3  will  be  found  there :...'.:  m  '.  n  :  ...  \ 

therefore  the  number  of  times  a  will  be  found  there 

m  1.2...(w  +  w  + ...)  1.2...(7n  +  »i+...-l) 

—  =  m 


??i +  «  +  ...  [1.2... m)  [1.2... n)...  (1.2...w)  (1.2...w)...  ' 

Similar  expressions  holding  for  the  number  of  times  y3,  7,...  ■will  ■ 
be  found  there :  we  have,  if  S  be  the  sum  of  the  digits  in  any 
assigned  place, 

[ma  +  n^+  ...)  \.2...[m  +  n-\- ...  -  1)  _ 
[1.2... m)  [1.2... n)...  ' 

therefore  if  2  denote  the  sum  of  all  the  numbers, 

2  =  ;S(1  +  10  +  10'''  +  ...  +  iO"'+"+--^), 

=  ^ 9 ' 

_  10»'+"^--_l  (»;a  +  7?/3+...)  1.2...(7>? +  »+•..-!) 
~  9  [1.2... m)  [1.2... n)...  ' 

the  sum  required. 

3.  The  difference  between  the  arithmetic  and  geometric 
means  of  two  niunbers  is  less  than  one-eighth  of  the  squared 
difference  of  the  numbers  divided  by  the  less  number,  but 
greater  than  one-eighth  of  such  squared  difference  divided  by 
the  greater  nimaber.  If  a*,  y  be  any  two  nimibers,  a-^,  y,  their 
arithmetic  and  geometric  means,  a:^,  y^  the  anthmetic  and  geo- 
metric means  of  iCj,  3/,,  and  so  on,  find  major  and  minor  limits 
for  the  difference  ic„  -  ?/„. 


16  SOLUTIONS   OF   SENATE-HOUSE    PKOliLEMS.  [iHol. 

(a)    Taking  the  notation  of  the  Litter  part  of  the  problem, 
we  have 

x-2  (a-M)i  +  y       (x*  -  iAY 
•  •    ^,      ^1  -  2  ~  2         * 

Now,  one-eighth  of  the  squared  difference  of  the  numbers 


K-yJ 


8 


4 

And  [x^-\-y'^y  lies  between  [2y^f  and  (2a;*)'\ 
or   >  4?/,     <  4a? ; 
4 


> 


3/        8       ' 

1    [x-yY 


X        8 
(/3)    From  above  it  appears  that 

„  _  „,   ^  (^«-i  ~  3/«-i)        ^  (^«-i  ~  3/»-i)  . 
therefore,  a  fortiori^ 


X    —  V    <      ^       (^"-2      ^/n-a) 


2  ' 

<    > 


< 


{^-yf 


1 

(^«-2        3/«-2) 

8^«-, 

.            (8^.-2)^          ' 

(^-2/f 

%„-.(83/„-2)^"(%)''"'        8a.,^_,(8.0^..(8a.)^"-^ 


4.  If  all  the  sums  of  two  letters  that  can  be  formed  with 
any  n  letters  be  multiplied  together,  then  in  each  term  of  the 
product,  the  sum  of  any  r  of  the  indices  cannot  exceed  the 
number  rn  —  \r[r-\-\). 


1851.]  ALGEBRA.  17 

Let  rtj,  a.^, f/^^,  be  the  letters. 

Then  the  product  will  be  of  the  form 

Now,  any  one  of  the  letters,  as  «|,  only  appears  n—\  times  in 
this  product,  that  is,  once  in  each  of  the  factors  a^  +  a.^ . . .  a,  +  r/,^, 
therefore  in  no  term  can  its  index  be  greater  than  n—\. 

And  in  such  terra,  the  index  of  a^  cannot  be  greater  than 
w— 2,  for  a,^  can  only  enter  n—2  times  as  a  factor  of  such  a  tenn, 
the  factor  a,  +  a.^  being  excluded. 

Similarly,  in  the  term  involving 

a"~^a^~'^  the  index  of  a^  cannot  be  >  7?  —  3, 


>i-l      n-2  n-r+l 

ttj     rtj     ...(i,_^        a,. >  ?«  —  r, 

therefore  the  sum  of  any  r  of  the  indices  cannot  be  greater  than 

(m-1)  +  (n-2)  +...  +  [n-r), 

=  rn  -  -^r  (/•+!), 
the  required  limit,  which  the  sum  of  the  r  indices  cannot  exceed. 

5.    Eliminate  x  from  the  equations 

[x  —  a)  [x  —  h)  =  [x  —  c)  (x  —  d)  =  [x  —  e)  [x  —f)^ 

and  from  the  same  equations  with  the  additional  relation  e  =/", 
find  a  quadratic  equation  for  determining  the  quantity  e  or  f. 
Shew  also  that  if  ?«',  vi"  be  the  values  of  e  or  f^  then  m"  —  m  is 
a  harmonic  mean  between  a  —  in\  b  —  m\  and  between  c  —  m\ 

d  —  m. 

(a)    Since  [x  —  a]  [x  —  h)  =  [x  —  c)  [x  —  d)  =  [x  —  e)  [x—f), 
we  get 
x^  —  [a  +  b)  X  +  ah  =  x'^  —  [c  +  d)  x  +  cd  =  x^  —  {e  +f)  x  +  ef] 

.'.    ie  +f—  a  —  b)  X  =  ef  —  ab^ 
[e  +f  —  c  —  d)  X  =  ef  —  cd^ 
...  [ef-  cd)  [e  +f-  a-b)  =  [ef-  ab)  {e  +/'-  r  -  d). ..{!), 


18  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

an  equation  from  which  x  is  eliminated,  and  which  may  be  put 
in  the  more  symmetrical  form 

ab{c  +  d-e-f)+ccl[e+f-a-h)  +  ef{a  +  h-c-(l)=0...{2). 

(/9)    If  ('  =/,  equation  (1)  becomes 

[e'  -  cd)  {2e  -  a  -  b)  =  {e'  -ah){2e-c-  d), 
...  i^c-\-d-a-h)  i  +  2[ah  -  cd)  e -\-  {a  +  b)  cd  -  [c -\-  d)  ab  =  0..  .(3), 
the  quadratic  for  the  deteniiination  of  e  or/. 
(7)    If  m'^  m"  be  the  roots  of  equation  (3), 

ab  —  cd 


m  +  m"  =  2 


mm 


a  +  b  —  c  —  d'' 

ab  {c  +  d)  —  cd  [a  +  b)  ^ 
a  +  b  —  c  —  d        ' 

{ab  —  cd)  {a  +  b) 


.'.    2  (ah  +  m'm")  =2  ,  ,  , 

^  '  a  +  b - c -d  ^ 

=  (a  +  b)  [m  +  m"). 
Now 
2 {ab->rm'm')  —  {a  +  b)  {m'  +  m")  =  [b—m') {a—m")  —  {a  —  m') {m"—  b), 

whence  {a  —  on)  {m"  —  b)  =  {b  —  m)  {a  —  m")^ 

.'.  a  —  m'  '.  b  —  m!  '.'.  a  —  m"  :  m"  —  b^ 

or   a  —  m'  :  b  —  m  ::  a  —  m  —  {m"  —  m')  :  m"  —  m  —  {b  —  m')^ 

whence  m"  —  in'  is  a  harmonic  mean  between  a  —  m\  b  —  on. 
Similarly,  it  is  a  harmonic  mean  between  c  —  on,  d  —  on. 


19 


TRIGONOMETRY. 

1848. 

1.  The  angles  of  a  quadrilateral  inscribed  in  a  circle  taken 
in  order,  when  multiplied  by  1,  2,  2,  3,  respectively,  are  in 
Arithmetical  Progression;  find  their  values. 

Let  6^  (j)  be  two  adjacent  angles  of  the  quadrilateral,  then 
TT  —  6,  TT  —  (j),  will  be  the  angles  respectively  opposite  to  them ; 
and,  by  the  conditions  of  the  problem,  0,  2^,  2(7r  —  6),  3(7r  —  0 1, 
are  in  Arithmetical  Progression. 

.-.    2(f)-  e  =  2(7r-^)  -  2(f)  =  S[7r-(f))  -  2(7r-^), 
.-.    A(f)  +  e  =  27r, 
4^  -  (/)  =  TT, 
.-.    17^=  Gtt, 
17<^  =  Ttt, 

17  '      ^       17  ' 
a       llTT  ,        IOtt 

17    '  ^        17    ' 

the  required  values  of  the  angles  of  the  quadrilateral. 

2.  Prove  that  sin 3^  sin'^  +  cos3^  cos'^  =  cos^2^. 
We  have  cos  3^  =  4  cos"^  —  3  cos  ^, 

sin  3^  =  3  sin^  -  4  sin'^; 
.-.  sin 3^  sin'^  +  cos 3^  cos'^ 

=  3  (sin*^  -  cos'<9)  +  4  (cos«^  -  sin"^), 
=  3  (cos^^  +  sin''^)  (sin'(9  -  cos*^)  +  4  (cos«^  -  sin"^), 
=  cos'^  -  3  cos'^  sin'-'^  +  3  cos'^  siu'(9  -  sin"!?, 
=  (cos'^-sm''^)=', 
=  cos"  2^, 
the  required  result. 

C2 


20  SOLUTIONS   OF  SENATE-HOUSE   PKOBLEMS.  [1848. 

3.  Having  given  the  three  right  lines  di'awn  from  any  point 
to  the  three  angular  points  of  an  equilateral  triangle,  determine 
a  side  of  the  triangle. 

Let  ABC  (fig.  16)  be  the  triangle,  0  the  point  from  which 
the  lines  are  dra^vn,  OA  =  a,  OB  =5,  0  0  =  c ;  also  let  the 
angle  BA  0=0^  CA  0  =  (f>^  and  let  a  side  of  the  triangle  =  x. 
Then,  by  the  triangle  BA  0, 

x'  +  d^  -  2ax  cos6  =  If (1). 

By  the  triangle  CA  0, 

x^  +  d^  —  2ax  cos(f)  =  c^ (2). 

Also  6  +  (f)  =  ^TT. 

Adding   (1)   and  (2),  we  get,  observing  that  cos  ^  + cos  ^ 
^       0-\-  cf>       e-<f> 

=  2  cos  — — ^  cos —!-  , 

2(a;^  +  a')  -  ^ax  cos^tt  cos  — -^  =  W  +  c' (3). 

Subtracting    (2)   from    (1),    and   observing    that    cos  ^  — cos  ^ 
=  2  sm  —  -     sm  — ^ — -  , 

Q  JL 

Aax  sin^TT  sin — — -^  =  b^  —  c^ (4). 


By  (3)  and  (4), 

=^  ^^ 

COS^  ^TT  sin""'  ^TT 


,,,'^  ji+i^\'^-:±^  ^ac^ff , 


...   [y^  +  c^  _  2d'y  -  ix\W  +  6'  -  2d'  +  ?>a^)  -h  3  (//  -  d')'  +  ^x'  =  0, 
...   a;*  _  [a'  +  P  +  c')  x'  +  a'  +  b*  +  c*-  {bV  +  c'd'  +  d'F)  =  0, 

2         d'  +  b^  +  d'  {{d'+b^+Cy  ,     4    ,     ,4     ,        4N     ,     72    2,       2     2,        271! 

.-.  x'= ±  <  ^ —  (a* -I-  b*  +  c)  +  Fc+d'ar^-a'b' 

=  '"'"^t'"^'''  ±  %  [2  (?>V  +  cV  +  d'h')  -  [a'  +  b'  +  c*)}4, 
which  determines  a  side  of  the  triangle. 


1849.]  TRIGONOMETRY.  21 

1849. 

1.    If  0  —  a,  0,  0  +  a,  be  three  angles  whose  cosines  are  in 
Harmonical  Progression,  prove  that 

cos  (f>  =  2^  cos  ^a. 

Since  cos(^  — a),  cos</),  cos(^  +  a)  are  in  Harmonical  Pro- 
gression, we  have 

2  1  1 

+ 


cos  <f>      cos  {(}>  +  a)      cos  ((jb  —  a) ' 

2  cos<^  cos  a 

cos  (0  —  a)  cos  (^  +  a)  ' 

/.    cos^<^  cosa  =  cos(^  — a)  cos((^  +  a), 

=  ^  (cos  2(f)  +  cos  2a), 

=  cos^^  —  sin'^a, 

2 ,  sin'^'a 

.*.    cos  ©  = , 

1  —  cosa 

4  sin'"*  ^a  cos'"*  ^a 
~        2  sin'  ^a        ' 

=  2  cos"''^a, 

.*.    cos<^  =  2-  cos^a, 

the  required  relation. 

2.  A  person  wishing  to  ascertain  his  distance  from  an  in- 
accessible object,  finds  three  points  in  the  horizontal  plane  at 
which  the  angular  elevation  of  the  summit  of  the  object  is  the 
same.     Shew  how  the  distance  may  be  found. 

Let  0  (fig.  17)  be  the  foot  of  the  object;  A^  B^  C  the  three 
points  at  which  the  angular  elevation  of  the  summit  of  the 
object  is  the  same ;  then  they  must  all  be  at  the  same  distance 
from  0.     Let  x  be  this  common  distance. 

Let  the  angle  AOB  =  d^  the  angle  AOC  =  <f).  Measm-e 
-BO,  CA^  AB,  and  let   their   distances  =  r/,  b,  o,   respectively ; 


22  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

then  a  =  2x  sin  ^[d  +  ^), 

b  =  2x  sin  ^<^, 

c  =  2x  sin  ^0. 
Eliminate  ^,  <^,  from  these  three  equations,  then  x  will  be 
known   and   the  distanee  of  the  person  from  the  object  deter- 
mined. 

1850. 

A  person  wishing  to  ascertain  the  distances  between  three 
inaccessible  objects  -4,  ^,  (7,  (fig.  18),  places  himself  in  a  line 
with  A  and  7i;  he  then  measures  the  distances  along  which  he 
must  walk  in  a  direction  at  right  angles  to  AB^  until  A^  (7,  and 
-B,  0,  respectively,  are  in  a  line  with  him,  and  also  observes  in 
those  positions  their  angular  bearings:  shew  how  he  can  find 
the  distances  between  A^  B,  and  C. 

Let  BE  and  BF,  the  measured  distances,  =  d  and  e ;  BEA 
and  BFA  the  obsei-ved  angles  =  a  and  /3.  Let  the  sides  of  the 
triangle  ABC  =  a,  h,  c,  and  BB  =  x. 

Therefore     tan  BEA  =  ^4^  , 

a     ' 

and  tan  BEB  =  ^  , 
a 

.-.  tan  a  =  tRn{BEA  -  BEB), 

X  +  C         X 

d         d 


^^x[x+cy 


d' 

cd 


d'^  +  x{x-{-  c) 


(1). 


Similarly,  tan^S  =  ~ —, ^ f2). 

e-  +  x{x  +  c)  ^  ^ 

From  equations  (1)  and  (2),  a;  and  c  are  known,  and  thence 

BAC  =  tan-^  -^  ,     and   CBA  =  180°  -  tan"^  -  , 
and  thence  the  distances  a  and  h. 


1851.]  TRIGONOMETRY.  2^ 

1851. 

.     T<>  /^      w  sin  a  cos  a 

1.    It  tanp  = ;— rr-  , 

1  —  /i  sin  a 

shew  that  tan(a-/3)  =  (1  — w)  tana. 

■iiT    1  /        ^N        tana  —  tan/3 

We  have    tan  (a  -  /3)  =  :; : : — ^  , 

^       ^'       I  +  tana  tan/3  ' 

71  sin  a  cos  a 

tan  a  — ; ;— 5— 

1  —  n  sm  a 

??  sin'' a 
1  + 


1  —  n  sni  a 


sin  a 

=  tan  a  —  n n  sm  a  cos  a, 

cos  a 


sin  a  —  «(sin  a  +  cos  a)  sin  a 


cos  a 
=  (1  —  w)  tan  a. 

2.  Two  triangles  stand  on  the  same  base,  determine  in  terms 
of  the  base  and  of  the  tangents  of  the  angles  at  the  base,  the 
distance  between  the  vertices  of  the  triangles. 

Let  ABG^  ABC  (fig.  19)  be  the  two  triangles.     Let  BC  the 
base  =  rt,  and  let  the  angles  ABC^  A  CB  =  B^  C,  respectively, 
and   the    angles   A'BC,  A'  CB  =  j5',  C      Join  AA',   and   let 
AA'  =  ?•,  then  it  is  required  to  find  the  magnitude  of  r. 
Draw  ABj  AD'  pei-pendicular  to  the  base.     Then 
r'=  [AD-AD'f  +  DD'% 

=  {AD  -  AD'f  +  {AD  cot  B  -  AD  cot  By. 
Now  a  =  AD  (cot  B  +  cot  (7), 

also  =  AD  (cot  B'  +  cotC) ; 

1  1 

r'  =  a' 


cotB  +  cotC      cot  5'  +  cotO'; 

cot-B'  cot  5 


,  /        coti^' coti^        Y 

"^  "  Vcot  B'  +  cot  C      cot  B  +  cot  67 


24  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1851. 

( /  tan  ^  tan  C  tan  B'  tan  C" 

(Vtan5  +  tanC      tani?  +  tanO 

tanC  tanC       N^)* 


,tani?'  +  tanC"      tan  5  +  tan  (7/  j  ' 
an  expression  of  the  required  form. 


25 


CONIC  SECTIONS. 

1848. 

1.  Given  the  lengths  of  the  axes  of  an  ellipse,  and  the 
positions  of  one  focus,  and  of  one  point  in  the  curve :  give  a 
geometrical  construction  for  finding  the  centre. 

Let  MN  (fig.  20)  be  a  line  equal  in  length  to  the  axis-minor. 
With  N  as  centre  and  a  radius  equal  to  the  axis-major,  describe 
an  arc  of  a  circle.  From  31  draw  MO  perpendicular  to  MN^ 
and  cutting  the  arc  in  0,  MO  will  be  equal  to  the  distance 
between  the  foci  of  the  ellipse. 

Produce  8P  to  Q  (fig.  21)  making  ;S^^  equal  to  NO.  With 
P  as  centre,  and  PQ  as  radius,  describe  an  arc  of  a  circle,  and 
Avith  S  as  centre,  and  radius  equal  to  MO^  describe  another  arc ; 
H  the  point  of  intersection  of  these  arcs  will  be  the  other  focus, 
for  /SlP,  PH  are  together  equal  to  the  axis-major,  and  SII  is 
equal  to  the  distance  between  the  foci.  If  therefore  we  bisect 
SH  in  (7,  C  will  be  the  centre. 

Since  the  arcs  described  from  6^,  P  as  centres  will  in  general 
intersect  in  two  points,  it  appears  that  there  are  two  positions 
which  the  centre  may  have. 

2.  P  is  any  point  in  an  ellipse  (fig.  22),  A  A'  its  axis-major, 
NP  an  ordinate  to  the  point  P;  to  any  point  Q  in  the  curve 
draw  AQ^  A'  Q^  meeting  NP\n  R  and  S-^  shew  that 

NR.NS  =  NP\ 

Draw  the  ordinate  QM,  then  by  similar  triangles  ANP^  AMQ, 

XR  :  NA  ::  MQ  :  MA, 

an<l  by  similar  triangles  A'NS,  A'MQ, 

NS:  NA'  ::  MQ  :  MA\ 


26  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

therefore        NE.NS :  NA.NA  ::  MQ"  :  MA.MA  , 

also  NP^  :  NA.NA'  : :  BC  :  A  G\ 

therefore  NR.N8  =  NP\ 

3.  FSp  (fig.  23)  is  a  focal  chord  of  a  parabola,  BDr  the 
directrix  meeting  the  axis  in  2) ;  Q  is  any  point  in  the  curve : 
prove  that  if  PQ,  p  Q  produced  meet  the  directrix  in  i?,  r,  half 
the  latus-rectum  will  be  a  mean  proportional  between  DR.,  Dr. 

Draw  PjV,  Qm  pei'pendicular  to  the  directrix,  and  join 
/ST?,  Sr^  SQ. 

Then  sinPi^^S'  =  sin  P8R  ^ 

PN 
=  dnPSR  ^ 

=  smPSR.&mPRN  : 
similarly  sin  QR8  =  sin  QSR.sin  QRNj 

.-.   smP8R  =  s'ln  Q8R, 
.:    Q8R=p8R: 
similarly  Q8r  =  P8r, 

.'.    R8r  is  a  right  angle, 

.-.    8D'  =  DR.Dr, 

or  half  the  latus-rectiun  is  a  mean  proportional  between  DR.,  Dr. 

1849. 

1.  Draw  a  parabola  to  touch  a  given  circle  in  a  given  point, 
so  that  its  axis  may  touch  the  same  circle  in  another  given 
point. 

Let  PQR  (fig.  24)  be  the  given  circle,  P  the  point  in  which 
the  parabola  is  to  touch  it,  Q  the  point  in  which  the  axis  is  to 
touch  it.  Draw  PT  a  tangent  to  the  circle  at  P,  this  will  also 
be  a  tangent  to  the  parabola  at  P.  Draw  QT  touching  the 
circle  at  (),  and  at  the  point  P  in  the  straight  line  PP,  make  the 


1849.]  CONIC   SECTIONS.  27 

angle  TP8  equal  to  the  angle  PTQ]  then  S^  the  intersection 
of  QT  and  FS,  will  be  the  focus  of  the  parabola.  Bisect  TF  in 
K,  and  draw  KA  perpendicular  to  ST.  A  will  be  the  vertex 
of  the  parabola,  and  the  vertex  and  focus  being  found,  the  cui^ve 
may  be  constructed. 

2.  If  a  circle  be  described  touching  the  axis-major  of  an 
ellipse  in  one  of  the  foci,  and  passing  through  one  extremity 
of  the  axis-minor,  the  scmiaxis-major  will  be  a  mean  propor- 
tional between  the  diameter  of  this  circle  and  the  semiaxis- 
minor. 

Let  AA'  (fig.  25)  be  the  axis-major  of  the  ellipse,  S  the 
focus,  C  the  centre,  B  the  extremity  of  the  axis-minor.  De- 
scribe the  circle  S£P  touching  AA'  in  >S',  and  passing  through 
By  and  draw  the  diameter  SB. 

Join  SBy  BB;  then  the  angle  SBB^  being  in  a  semicircle,  is 
a  right  angle,  also  the  angle  SCB  is  a  right  angle.  And  the 
angle  BSP  is  equal  to  the  angle  SBCj  therefore  the  triangles 
BSBy  SBC  are  similar.     Hence 

BC:  SB::  SB:  SB, 

or  SB  is  a  mean  proportional  between  SB  and  BC.  But  SB 
is  equal  to  the  semiaxis-major ;  therefore  the  semiaxis-major 
is  a  mean  proportional  between  the  diameter  of  the  circle  and 
the  semiaxis-minor. 

3.  If  ABj  CBj  two  lines  in  an  ellipse,  not  parallel  to  one 
another,  make  equal  angles  with  either  axis ;  the  lines  A  C,  BD 
and  ADy  BC  will  also  make  equal  angles  with  either  axis. 

Let  A'B'C'D'  (fig.  26)  be  the  points  of  intersection  of  the 
peq^endiculars  to  the  axis-major  through  the  points  ABCD  with 
the  auxiliaiy  circle  A'B'C'D'.  Then  it  is  evident  that  a  line 
joining  any  two  of  the  above  points  as  A'B'  will  intersect  the 
axis-major  in  the  same  point  as  AB  does,  and  any  two  lines 
joining  the  above  points  as  A'B'.^  CD'  will  be  equally  inclined 
to  the  axis-major,  and  therefore  to  either  axis,  if  AB.,  CD  arc  so. 


28  SOLUTloX.S   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

and  vice  versa:  hence  we  have  to  prove  that  If  A'B\  CD'  are 
equally  inclmed  to  the  axis-major  LFE^  the  lines  A'C'^B'D' 
and  A'D\  B'  C  are  so. 

Now  lAGL  =  L  A  EL  +  L  B'A  C\ 

and  z  D'HL  =  l  UFL  +  l  B'D'  C  ; 

also  /:AEL  =  I  UFL, 

and  lB'AC  =  aB'D'C; 

therefore  lAGL  =  L  D'HL  =  L  B'HG, 

or  AC\  B'D'  are  equally  inclined  to  LFE. 

Again,  z  ALH  =  L  LD'H  +  z  LED', 

and  z B'KG  =  lKC G  ^  lKGC ', 

also  lLD'H=lKC'G, 

and  lLED  =  lKGC] 

therefore  z  ALE  =  L  B'KG, 

or  AD'  and  5 '6"  are  also  equally  inclined  to  LFE\  therefore 
also  AC,  BD  and  AD,  BC  are  equally  inclined  to  either  axis. 

Q.  E.  D. 

1850. 

1.  If  from  any  point  P  of  a  circle,  PC  be  drawn  to  the 
centre  (7,  and  a  chord  PQ  be  drawn  parallel  to  the  diameter 
A  CB,  and  bisected  in  R,  shew  that  the  locus  of  the  intersection 
of  CP  and  AR  is  a  parabola. 

Let  0  (fig.  27)  be  the  intersection  of  CP  and  AR.  Draw 
AM,  CN  pei-pendicular  to  AB,  ONM  parallel  to  AB.  CN  will 
pass  through  R.     Then 

CO-.CP'.'.  AO:AR, 

::  OM:MN. 

But  CP=AC=MN, 

therefore  CO  =  OM, 

and  the  locus  of  0  is  a  parabola,  of  which  C  is  the  focus,  AM 
the  directrix. 


1850.]  CONIC  SECTIONS.  29 

2.  From  the  point  P  in  the  ellipse  APB  (fig.  28),  lines  are 
drawn  to  A^  B,  the  extremities  of  the  axis-major,  and  from 
A,  B^  Hues  arc  drawn  perpendicular  to  AP^  BP;  shew  that  the 
locus  of  their  intersection  will  be  another  ellipse,  and  find  its 
axes. 

Let  Q  be  the  intersection  of  the  lines  pei'pendicular  to 
AP^  BP.  Draw  P2f,  QN  perpendicular  to  the  axis-major,  then 
the  triangles  PBM,  BQN  ^yl\\  be  similar,  therefore 

PM:BM::BN:  QN', 
similarly  P3I :  AM ::  AN :  QN, 

therefore  PM'  :  AM.BM : :  AN.BN :  ^.V''. 

But  if  Z*  (7  be  the  semiaxis-minor  of  the  original  ellipse, 

PM'  :  AM.BM::  bC  :  AC\ 
therefore  AN.BN :  QN'  ::hC'' :  AC] 

therefore  the  locus  of  Q  is  an  ellipse  whose  axes  are  to  one 
another  as  IC :  AC. 

And  if  we  draw  AB\  BB'  perpendieidar  to  Ah^  Bb^  we  have 

A  (1^ 

^■^=#' 

bC 
which  is  one  axis,  the  other  is  equal  to  -^-^  B'C  or  A  C. 

3.  If  two  elUpses  having  the  same  major  axes,  can  be 
inscribed  in  a  parallelogram,  the  foci  of  the  ellipses  will  lie 
in  the  comers  of  an  equiangular  parallelogram. 

For  it  is  evident  that  the  centres  of  the  ellipses  must  lie  at  the 
point  of  intersection  of  the  diagonals  of  the  parallelogram,  that 
is,  must  be  coincident,  and  their  major  axes  are  equal;  there- 
fore they  will  have  a  common  auxiliary  circle. 

The  lines  joining  the  points  of  intersection  of  this  circle  and 
the  parallelogram,  will,  if  the  right  points  are  joined,  be  pei'pen- 
dicular to  the  sides  of  the  parallelogram,  and  each  of  them  will 
contain  two  foci:  hence  the  four  foci  will  be  at  their  points  of 
intersection,  that  is,  at  the  comers  of  a  parallelogram,  equiangular 
with  the  circumscribing  parallelogram. 


30  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [IBol- 

4.  If  from  the  extremities  of  any  diameter  AB  (fig.  29)  of 
an  equilateral  liyperbola,  lines  be  drawn  to  any  point  P  in  the 
curve,  they  will  be  equally  inclined  to  the  asymptotes. 

From  A  and  P  draw  the  perpendiculars  A  C,  AF,  PE,  PF, 
on  the  asymptotes,  AF  and  PF  intersecting  in  Fj  from  B  draw 
BD  perpendicular  to  the  asjTuptote  Oq.  Then,  since  PQ  =  Bq., 
and  the  triangles  QFP^  BDq  are  similar,  they  are  also  equal ; 
therefore  PE  =  Dq  and   QE  =  BD,  therefore 

AF=^  CO  +  EP=OD  +  Dq  =  Oq, 

and      PF  =  OE  +  AC  =  OE  +  BB  =  OE -\-  EQ  =  OQ', 

therefore  the  right-angled  triangles  APF,  qQO  are  equal  in 
all  respects,  and  the  chords  PA,  PB  equally  inclined  to  the 
asymptotes. 

1851. 

1.  Given  a  pair  of  conjugate  diameters  of  a  conic  section, 
find  geometrically  the  position  of  the  principle  diameters,  (1) 
in  the  case  of  the  hyperbola,  (2)  in  that  of  the  ellipse. 

Let  PP',  DU  (fig.  30)  be  the  given  conjugate  diameters 
of  an  hj-perbola  intersecting  in  C. 

Join  PD,  PB'-,  bisect  them  in  E  and  F,  draw  GE,  OF', 
bisect  the  angle  ECF  by  the  line  A'  CA,  and  through  C  draw 
BCB'  pei'peudicular  to  ACA' ',  these  will  be  the  principal  di- 
ameters required.  For,  by  the  property  of  the  hyperbola, 
CE,  CF  are  the  asjnnptotes,  and  AC  A',  BCB',  to  which  they 
are  equaUy  inclined,  are  the  principal  diameters. 

2.*  The  solution  of  this  part  of  the  problem  depends  upon 
the  property  of  the  ellipse,  that  if  P  (fig.  31)  be  any  point  in  the 
ellipse,  and  CR,  CN,  two  lines  at  right  angles  to  each  other,  cut 
the  straight  line  PRN  in  the  pomts  R,  N,  such  that  PN  is  equal 
to  the  semiaxis-major,  and  PR  to  the  semiaxis-minor,  CR  and 
CN  will  be  the  directions  of  the  principal  axes. 


*  For  this  solution  the  authors  are  indebted  to  the  kindness  of  the  Mode- 
rator, Mr.  Gaskin. 


1851.]  CONIC   SECTIONS.  31 

Let  CP,  CD  be  the  given  semi-conjugate  diameters;  draw 
Pi^  pei-pcndlcular  to  CD'^  make  FK  equal  to  CP;  upon  CK  as 
diameter  describe  the  circle  CFK\  tlii'ough  its  centre  0  draw 
PRN'^  join  CP,  CN'.  these  will  be  the  directions  of  the  principal 
axes. 

For         PF.  CD  =  PF.PK  =  PR.PN  =  A  C.BC, 

and  CP'  +  CD'  =  CP''  +  PK'  =  2  CO'  +  2P0'  =  2  OE'  +  2  0P\ 

=  PR  +  P^  {Euc.  II.  10)  =  J.0"^  +  PC'^ 
therefore  PN  =  AC,     PR  =  BC, 

and  OP,  CW  are  the  du*ections  of  the  principal  axes. 


{     32     ) 


STATICS. 

1849. 

1.  Two  forces  F  and  F'^  acting  In  the  diagonals  of  a  paral- 
lelogram, keep  it  at  rest  in  such  a  position  that  one  of  its  edges 
is  horizontal;  shew  that  i^seca  =  i^'seca' =  TFcosec(a  +  a'), 
where  W  is  the  weight  of  the  parallelogram,  a  and  a!  the  angles 
between  its  diagonals  and  the  horizontal  side. 

Let  AB  (fig.  32)  be  the  horizontal  side  of  the  parallelogram. 
In  order  to  preserve  equilibrium,  the  directions  of  the  forces 
F^  F'  must  meet  in  G^  the  centre  of  gravity.  Hence,  by  the 
triangle  of  force, 

F       _       F'       _       W 
smBG  W ~  &\nA G  W ~  sin^ GB ' 
F  F'  W 


or 


cosa  cosa  sm(a  +  a) 

therefore  i^seca  =    i^'seca'    =  TFcosec(a -I-  a'). 

2.  A  cubical  box  is  half-filled  with  water,  and  placed  upon 
a  rough  rectangular  board ;  if  the  board  be  slowly  inclined  to  the 
horizon,  determine  whether  the  box  will  slide  dowTi  or  topple 
over. 

Let  fi  =  the  coefficient  of  friction. 

Then  the  box  would  begin  to  slide  when  the  inclination  of 
the  board  to  the  horizon  =  tan~^yLt. 

It  would  begin  to  topple  when  the  inclination  =  jtt. 

Therefore  it  will  begin  to  slide  or  topple  over,  according  as 
/Lt  <  or  >  1. 

1850. 

1,  A  heavy  body  is  supported  in  a  given  position  by  means 
of  a  string  which  is  fastened  to  two  given  points  in  the  body. 


1850.]  sTATifs.  •};^ 

and  then  passes  over  a  sjnooth   peg:    find  the   k-ngtli   ot"  the 
strmg. 

Let  G  (fig.  33)  be  the  given  position  of  the  centre  of  gravity, 
A  and  B  those  of  the  points  of  support.  The  position  of  tlie 
peg  P  is  determined  by  the  conditions  that  it  must  lie  in  the 
vertical  tlu'ough  G^  and  that  the  angles  APG,  BPG  must  bi; 
equal,  each  =  6  suppose. 

luQt  AG  =  a,  BG  =  b,  lA GP  =  a,  L BGP  =  /3 ;  then 
PG      miPAG      sinf^  +  a) 


AG      sin  APG  sin6> 


=  cos  a  +  sin  a  cot^ 


PG 
similarly  -^^  —  cosyS  +  sin^cot^, 

,       ^  cosa  +  sina  cot^       BG       b 

thereiore  y^ -. — ^ 7:  =  -j-f,  =  -  , 

cosp  +  smp  cota      AG      a 

whence  6  is  known,  and  length  of  the  string 

=  AP^BP, 

sin  a         sin  /3  ,  .    , 
=  -. — 7i  a  +  -. — -  b  is  known, 
sma         sma 

2.  Two  spheres  are  supported  by  strings  attached  to  a  given 
point,  and  rest  against  one  another:  find  the  tensions  of  the 
strings. 

Let  -4,  B^  (fig.  34)  be  the  centres  of  the  spheres,  and  C  the 
peg.  Then,  if  the  spheres  are  smooth,  the  strings  must  lie  in 
the  lines  CA^  CB;  hence  the  parts  of  the  triangle  ABC  ai-e 
known.     To  determine  its  position. 

Let  G  be  the  centre  of  gravity  of  the  spheres,  CG  must  be 
vertical.  Let  W^,  W^  be  the  weights  of  the  spheres  A  and  B, 
therefore  AG  :  BG  ::  W^  :  W^-, 

and  if  LACG  =  e, 

smjC-  6)  _  BG  sin^  _  TF,  sini?  ^ 
siiT^        ""  ATCf&mA  ~  it;  sin  ^  ' 

1        /•  •    />        /I  n       ^^^,  sin  5 

therefore  smC  cot  a  —  cosO  =  ttt   • — 7  1 

T^.^sm^  ' 

whence  6  is  known. 


34  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

Let  T,,  T^  be  the  tensions  of  the  strings  which  support 
A  and  B  respectively;  therefore  resolving  the  forces  on  the 
sphere  A  perpendicular  to  AB^ 

T,  sin^  -  PF;  sin(^  +  ^)  =  0, 

_  sin(^  +  ^)  ,.^  . 


or 


sin -4 


and  similarly  1\  =  t^^ ^  TT,, 

_sin(^  +  6>) 

~        sin  5  «' 

whence  T^  and  T^  are  known. 

3.  A  cone  of  given  weight  W  (fig.  35)  is  placed  with  its 
base  on  a  smooth  inclined  plane,  and  supported  by  a  weight 
W\  w^hich  hangs  by  a  string  fastened  to  the  vertex  of  the 
cone,  and  passing  over  a  pully  in  the  inclined  plane  at  the 
same  height  as  the  vertex.  Find  the  angle  of  the  cone  when 
the  ratio  of  the  weights  is  such  that  a  small  mcrease  of  W  would 
cause  the  cone  to  turn  about  the  highest  point  of  the  base,  as 
well  as  slide. 

Let  a  =  the  angle  of  the  plane, 

6  =  the  half-angle  of  the  cone. 
Since  the  resolved  parts,  along  the  plane,  of  the  tension  W  of 
the  string  and  the  weight  W  just  balance,  we  have 

PTsina  =  T'T'cosa (1) ; 

and'  because  the  moments  of  the  same  forces  about  B  are  also 
equal, 

Wmia.lAC+  Wcosa.BC  =  W  cosa.AC  -  W  sma.BC, 

W  cosa.f^O  =  (  W  ^  +  W  sin  a")  BC,  from  (1), 

or  BC  =  f  sina  cosa  A  C, 

or  tan^  =  #  sin  2a. 


1851.]  STATICS.  35 

1851. 

1.   A  cone  whose  semi-vertical  angle  is  tan"*  -j  is  enclosed 

in  the  circumscribing  spherical  surface,  shew  that  it  will  rest  in 
any  position. 

Let  ABC  (fig.  35)  represent  a  section  of  the  cone  made  by 
a  plane  through  its  axis.  Divide  the  axis  AD  in  (r,  so  that 
GD  =  iADj  then  G  will  be  the  centre  of  gravity  of  the  cone. 
Join  BGj  then 

BG'  =  BD"'  +  DG\ 


But 

i'AnBAD  =  ^^, 

therefore 

BD'  =  \AD\ 

and 

DG'  =  ^AD\ 

therefore 

BG'  =  ^AD\ 

and 

BG  =  IAD, 

^AG; 
therefore  G  is  the  centre  of  the  circiunscribing  sphere. 

Hence  it  appears  that  the  height  of  the  centre  of  gravity  of 
the  cone  will  be  the  same  in  whatever  position  it  be  placed, 
therefore  it  will  rest  in  any  position. 

2.  A  string  ABCDEP  (fig.  37)  is  attached  to  the  centre 
A^  of  a  pully  Avhose  radius  is  ?•,  it  then  passes  over  a  fixed  point 
B^  and  under  the  pully,  which  it  touches  in  the  points  C  and  D  ; 
it  afterwards  passes  over  a  fixed  point  E^  and  has  a  weight  P 

attached  to  its  extremity ;  BE  is  horizontal  and  =  — ,  and  DE 

is  vertical :  shew  that  if  the  system  be  in  equilibrimn  the  weiglit 

bP 

of  the  pully  is  — ,  and  find  the  distance  AB. 

Let  W  be  the  weight  of  the  pully,  and  let  ^,  ^  denote  the 
respective  inclinations  of  AB^  BC  to  the  horizon.  The  tension 
of  the  string  will  be  thi'oughout  =  P;  hence  resolving  horizontally 

d2 


36  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

and  vertically, 

Pcos^  -  Pcos<^  =  0  (1), 

P(l  +  sln^  +  sin</))  =  TF. (2). 

Again,  AB  =  r  cosec(^  +  ^), 

5?' 
therefore  EB  =  —  =  r  {I  +  cosec[6  +  </>)  cos^}, 

therefore  |sm(^  +  4>)  =  coaO (3). 

By  (1)  e  =  </,, 

and  by  (3)  fsm2^  =  cos^, 

therefore  sin^  =  ], 

5P 
therefore  by  (2)  W  =  —  . 

Also  AB  =  rcosec(^  +  ^), 

r 

^  sm2^  ' 

r 

-  -^ 

"3:7*' 

which  gives  the  distance  AB. 


(     37     ) 


DYNAMICS. 

1848. 

1.  Two  bodies  acted  on  by  gravity  are  projected  obliquely 
from  two  given  points  in  given  directions  and  with  given 
velocities:  determine  their  position  when  their  distance  is  the 
least  possible. 

Let  the  bodies  A^  B  be  projected  from  the  points  A^  B 
(fig.  38)  in  directions  AC^  BC  intersecting  in  (7,  and  with 
velocities  proportional  to  CE  and  CD]  upon  both  the  bodies 
impress  a  velocity  CE  equal  and  opposite  to  ^'s  velocity,  and 
suppose  gravity  not  to  act,  the  relative  motion  of  A  and  B  will 
not  be  affected  by  either  of  these  circumstances ;  but  A  will  now 
be  reduced  to  rest,  and  B  will  move  in  a  direction  BG  parallel 
to  the  diagonal  CF  of  the  parallelogram  on  CE^  CD.  From  A 
draw  AG  perpendicular  to  BG^  AG  will  be  the  shortest  possible 
distance  between  A  and  B]  and  A  and  B  will  be  at  that  dis- 
tance at  the  time  [t]  after  the  instant  of  projection  that  it  takes 
a  body  animated  with  the  velocity  CF  to  describe  the  space  BG^ 
a  kuo\\Ti  time  therefore.  Let  AH  and  BK  be  the  spaces  due  to 
^'s  and  -B's  velocity  of  projection  in  time  t.  Through  H  and  K 
draw  HL  and  KM^  each  equal  to  the  space  due  to  gravity  in  the 
tunc  t ;  L  and  M  are  the  positions  required. 

2.  A  railway  train  is  going  smoothly  along  a  curve  of 
500  yards'  radius  at  the  rate  of  30  miles  an  hour ;  find  at  what 
angle  a  plumb-line  hanging  in  one  of  the  carriages  will  be  in- 
clined to  the  vertical. 

Let  a  denote  the  inclination  of  the  plumb-line  to  the  vertical, 


38  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1848. 

CO  the  angular  velocity  of  tlie  train  per  second,  r  the  radius  of 
the  cur\'e. 

Then  the  weight  at  the  end  of  the  plumb-line  may  be  con- 
sidered to  be  in  oquilibrium  under  the  action  of  the  centrifugal 
force,  gravity,  and  the  tension  of  the  string. 

Hence  (fig.  39),  by  the  triangle  of  forces. 


sin  (tt  —  a)       sin  (^tt  +  a)  ' 
coS'  g 


or 


sm  a      cos  a 

tana  = . 

9 


AT  „.  .r^.  30x5280  44 

Now  g  =  32.2,  .  =  1500,  a>  =  ^^^^-^^^^^0  =  1500  ' 


(44 
.'.   tana  = 


1500  X  32.2  ' 

^   (44F 
48300  ' 

-    4^4 
~  12075  ' 

which  gives  the  inclination  to  the  vertical. 

3.   A  nmnber  of  balls  of  given  elasticity  A,  B,  C are 

placed  in  a  line ;    A  is  projected  with  a  given  velocity  so  as 
to  impinge  on  B]  B  then  impinges  on  (7,  and  so  on:  find  the 

masses  of  the  balls  B^  C ,  in  order  that  each  of  the  balls 

A,  B,   C may  be  at  rest  after  impinging  on  the   next; 

and  find  the  velocity  of  the  «*''  ball  after  its  impact  with  the 
(n  -  1)'". 

Let  m  :  1  be  the  ratio  of  the  mass  of  w*  ball  to  that  of  the 
{n—  1)'^',  then  the  ratio  of  the  velocity  of  the  {n—  ly^  ball  after 
impact  to  its  velocity  before,  would  be,  if  the  balls  were  inelastic, 

_       1 

~  1  +  m  ' 


1848.]  DYNAMICS.  39 

Since  they  are  elastic,  the  ratio  is 

=  1  -  (1+  .)  f  1  -       ^ 


=  1  -  (1  +  e) 


1  +  mj  ' 
m 


1  +  m 
and  since  the  (w  —  1)"*  ball  is  thus  brought  to  rest,  this  must  =  0, 

.-.    1  +  -  =  1  +  e, 
m 

and  ni  =  -  , 

e 

so  that  the  masses  of  the  balls  from  a  geometrical  progression, 
whose  common  ratio  is  -,  and  the  ratio  of  the  velocity  of  the 

I  -\-  e 
n*^  ball  after  impact  to  that  of  the  (n—  lY^  before  =  ;; =  e : 

therefore  if  V  be  the  initial  velocity  of  A,  velocity  of  n^^  ball 
after  impact  =  e"~'  V. 

4.  An  imperfectly  elastic  ball  is  projected  in  a  given  direction 
within  a  fixed  horizontal  hoop,  so  as  to  go  on  rebounding  from 
the  surface  of  the  hoop ;  find  the  limit  to  which  the  velocity  of 
the  ball  will  approach,  and  shew  that  it  will  attain  this  limit  at 
the  end  of  a  finite  time. 

Let  e  be  the  modulus  of  elasticity,  F,  V^ F^  the  velocities 

of  the  ball  before  the  first,  second, (n— 1)'^  impacts,  6,  ^,...^„ 

the  successive  angles  of  incidence.     Then 

Fj  cos  0^  =  e  Fcos  ^, 

F,8in^,  =  Fsin^, 

.-.    F/  =  F^(sin'^^  +  e"'cos^^), 
=  sin''^(l  +e:'coi'e)V': 
shuilarly    F/  =  sin'^6»,  {I  +  e'  cof''^,)  V;\ 
1  1 


But  sin'^.  = 


1  +cof^,       1  +  rVot'^' 


40  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1B49. 

therefore  T^  =  sin''6>  (1  +  c?  cof^^J  V% 

and  similarly  it  may  be  shewn  that 

T7  =  sin-^^(1  +  e'^"cof^^)F''; 
hence  when  n  is  indefinitely  increased, 

V  =  Fsin(9, 

the  limit  to  which  the  velocity  of  the  ball  approaches. 

Now  the  distances  between  the  successive  points  of  incidence 
are  2rcos^  ,  2?'Cos^.^ r  being  the  radius  of  the  circle;  there- 
fore the  times  of  describing  these  spaces  are 

2?-cos^,       2rcos^„  .     , 

— 1^- ' ,    — T^— ^ respectively, 

1  2 

cot  6',  1        ,  cot^„  1 


=  2r 


:i  +  cof^j*  TV      (i  +  cof^^j*  f; 

('cot^  (1  +  e^cot-^ji    1 


H-e'^cot'6')4  sin  (9  F' 

e'cot^          (l+e*cot'^)*    1 


2r 


:i  +e*cot'6')4  sin^  F"        ' 

2r     cos^       2r    jjcos^ 

""  F^ii^'     F^  dK^ ' 

therefore  the  ball  will  attain  its  terminal  velocity,  after  the  time 
2r  cos^   /  2        s         \ 

2r   cos^      e 


V  sin'^  I  -e' 

1849. 

1.  A  body  is  projected  from  a  given  point  in  a  horizontal 
direction  with  a  given  velocity,  and  moves  upon  an  inclined 
plane  passing  through  the  point.  If  the  inclination  of  the  plane 
vary,  find  the  locus  of  the  directrix  of  the  parabola  which  the 
body  describes. 

Let  a  be  the  inclination  of  the  plane  to  the  horizon ;  F  the 
velocity  of  projection ;   /  the  latus-rectum  of  the  parabola  de- 


1849.]  DYNAMICS.  41 

scribed ;  therefore 

j^r  sina  ' 

and  I  sill  a  = . 

9 

But  J?  sin  a  is  the  height  of  the  directrix  above  the  given 

point  of  projection ;  therefore  this  height  is  constant,  and  the 

locus  of  the  directrix  is  a  horizontal  plane  at  a  distance  —  above 

2g 

the  given  point. 

2.  An  imperfectly  elastic  ball  A  lies  on  a  billiard-table, 
deteniiine  the  direction  In  which  an  equal  ball  B  must  strike 
it  in  order  that  they  may  impinge  upon  a  side  of  the  table 
at  equal  given  angles. 

The  impact  must  be  oblique  and  the  impulse  take  place  in 
the  direction  in  which  A  is  to  go  off.  This  direction  makes 
with  the  side  of  the  table  the  given  angle  a :  let  ^  be  the  angle 
which  ^'s  direction  before  impact  makes  this  direction. 

V  =  B^s  velocity  before  impact, 

e  =  the  modulus  of  elasticity. 

5's  velocity  V  sin  0  pei'pendicular  to  the  direction  of  the  Impulse 
will  be  unaltered  by  it:  if  there  were  no  elasticity,  its  velocity 
in  direction  of  the  impulse  after  impact  would  be  ^Fcos^,  since 
the  balls  are  equal,  and  the  impulse  ^MJ^cosd:  hence  the  actual 
impulse  will  be  ^(1+e)  MVcosd,  and  the  actual  velocity  in  its 
direction  after  impact  P^cos^  —  ^{1  +  e)  Fcos  ^  or  ^(1  —  e)  Fcos^. 
Let  <f)  =  the  angle  which  ^'s  direction  after  impact  makes 
with  the  direction  of  the  impulse, 

tan(/)   =    yy- r ^   =    j— r  taU  ^. 

■^(l-e)cos^      i{l-(') 
But  (f>  =  2a, 
.-.  tan^  =  ^(1  -e)  tan 2a, 
whence  6  is  known. 


42  .SOLUTIUXS   OF   SENATE-HOUSE    PliOBLEMS,  [1849. 

3.  A  bead  running  upon  a  fine  thread,  the  extremities  of 
which  are  fixed,  describes  an  ellipse  in  a  plane  passing  through 
the  extremities,  under  the  action  of  no  external  force;  prove 
that  the  tension  of  the  thread  for  any  given  position  of  the  bead 
is  inversely  proportional  to  the  square  of  the  conjugate  diameter. 

Let  the  bead  be  at  the  point  P  of  the  ellipse. 

Since  the  tension  of  the  string  is  the  same  tkroughout,  the 

resultant  force   on  the  bead  will  bisect  the  angle   SPH^  and 

therefore  be  nonnal  to  the  elliptic  path.     Consequently,  as  no 

force  acts  upon  the   bead   in   the   direction  of  its  motion,  its 

velocity  will  be  uniform.    Now,  considering  the  bead  as  moving, 

for  the  instant,  in  the  circle  of  curvature  at  the  point  P,  nomial 

J,  vel.'  1  .,,... 

torce  Gc  — 3 — 7: cc  — j — ^ ,  smce  the  velocity  is  mii- 

rad.  ot  cui'v.       rad.  01  curv.  •' 

form :    but  radius  of  curvature  gc  CD\  therefore  normal  force 

1 

Now,  adopting  the  usual  notation, 
tension  of  the  string  :  nomial  force  -.-.PE:  PFr.  CD.AGxCD.PF, 

'.iGB-.BC, 

therefore  tension  of  the  string  cc  -^^  . 

4.  The  centres  of  two  equal  spheres  (elasticity  e,  radius  r,) 
move  in  opposite  directions  in  a  circle  (radius  B)  about  a  centre 
of  force  vaiying  inversely  as  the  square  of  the  distance ;  deter- 
mine the  motion  of  the  spheres  after  they  have  impinged,  sup- 

posing  that  e  =  ^ 5 ;  and  prove  that  the  latus-rectum  of  the 

conic  section  described  after  the  second  impact  will  be  2e^i?. 

Let  0  (fig.  40)  be  the  centre  of  force ;  (7,  C  the  centres  of 
the  spheres.  Draw  OPQ  perpendicular  to  CC\  such  that  CQ 
is  perpendicular  to  00,  and  consequently  C  Q  to  OC.  Then 
if  CQ  represent  in  magnitude  and  direction  the  velocity  of  the 
sphere  0  before  impact,  CPj  PQ  will  represent  its  resolved  parts 
in  directions  CP^  PQ.     Now  draw  QB  perpendicular  to  0'^, 


1850.]  DYNAMICS.  43 

meeting  CP  in  B:  the  triangle  QPR  is  evidently  similar  to 
G'PQ^  and  therefore  to  CPQ.     Hence 

BP:PQ::PQ:  CP, 

.:  PPiCP::  PQ'  :  CP'  ::  CP'  :  OP'  ::  r'  :  B'  -  r'  ::  e  :  I. 

Consequently  BP  represents  in  magnitude  and  direction  the 
resolved  part,  pei'pendicular  to  OQ,  of  (7's  velocity  after  impact. 
The  velocity  PQ  remains  unaltered  by  the  impact ;  therefore 
the  diagonal  of  the  parallelogram  PBQ  drawn  through  P  will 
represent  in  magnitude,  and  be  parallel  to  the  direction,  of  the 
whole  velocity  of  C  after  impact.  Now  this  diagonal  makes 
with  PQ  an  angle  equal  to  BQP  or  COP  or  COP,  and  is 
therefore  parallel  to  OC.  Hence  after  impact  the  centres  of 
the  spheres  will  move  directly  from  the  centre  0  in  the  lines 
OC,  OC.  They  will  evidently  return  to  the  same  positions 
C  and  C,  and  there  impinge  a  second  time. 

For  the  velocity  of  C  after  the  second  impact  it  is  sufficient 
to  obser\'e  that  the  velocity  along  OP  will  be  unchanged,  while 
that  perpendicular  to  OP  will  be  again  diminished  in  the  ratio 
of  ?:  1.  Let  PB'  =  e.PB.  Through  C  draw  CS  equal  and 
parallel  to  QB' ;  join  OS.  Therefore  the  latera-recta  of  the 
first  and  third  orbits  will  be  to  one  another  as  (triangle  OCQY 
:  (triangle  OSCf,  since  these  triangles  represent  upon  equal 
scales,  half  the  product  velocity  x  pei*pendicular  on  the  tangent ; 
and  we  may  shew  that  (triangle  OCQf :  (triangle  OSC)' ::  1 :  e* ; 
and  the  latus-rectum  of  the  first  or  circular  orbit  is  2B.  There- 
fore that  of  the  third  is  2e^B. 

1850. 

1.  Shew  that  it  is  possible  to  project  a  ball  on  a  smooth 
billiard-table  from  a  given  point  in  an  infinite  number  of 
directions,  so  as,  after  striking  all  the  sides  in  order  once  or 
oftener,  to  hit  another  given  point ;  but  that  this  number  is 
limited  if  it  have  to  return  to  the  point  from  which  it  was 
projected. 

Let  P  (fig.  41)  be  the  point  of  the  table  from  which  the  ball 
is  projected,  PQBSTU  its  course   once  round  the   table.     7?.S' 


44  S(.>LIT1U.NS   OF   SENATE-HUUSE   I'liUBLEMS.  [1850. 

may  be  sliewn  to  be  parallel  to  QP-^  and  if  the  elasticity  be 
perfect,  equidistant  with  it  from  the  line  AD  drawn  through  the 
comer  A  of  the  table  parallel  to  either  of  them.  For  the  angle 
SUB  =  angle  QRA  =  90°  -  BQA^^OT  -  PQD,  therefore  RS 
is  parallel  to  QP.     Also  if  QR  intersect  AD  in  F, 

RF'.QF::  RA  sin  RAF:  QA  smQAF, 

::  RA  s'mBRS  :  QA  sinDQP, 

::  RA  sm  ARF:  QA  sinAQF. 

But  RA  sm  ARF  =  QA  sin  A  QF, 

therefore  RF  =  QF^ 

and  RSj  QP  are  equidistant  from  AD. 

Similarly,  RS  and  TU  are  equidistant  from  CE.  Through 
P  draw  VDEPU  pei'pendicular  to  the  parallel  lines.  Then 
VD  =  DP  and  VE  =  EU,  therefore 

PU=  DE+  EU-  DP=DE+  EV -  DV=  2DE. 

The  same  equation,  PU  =  2DE,  holds  whether  P  and  U  be  on 
the  same  side  of  D  and  E  or  on  opposite  sides  of  either  or  both. 
Hence  it  is  evident  that  by  choosing  the  direction  PQ  rightly 
we  may  make  the  ball  hit  the  second  given  point,  through 
■which  the  line  TU  will  pass,  after  striliing  all  the  sides  once : 
and  by  lessening  DE  or  projecting  the  ball  more  nearly  in  the 
direction  of  the  diagonal  CA,  we  may  make  it  strike  the  second 
point  after  striking  all  the  sides  twice,  when  PU  will  =  ADEy 
and  so  on ;  there  being  thus  an  infinite  number  of  directions  of 
projection  each  more  nearly  parallel  to  the  diagonal  CA  than 
the  preceding,  which  will  cause  the  ball  to  hit  the  second  given 
point  after  striking  all  the  sides  once,  twice,  &c.,  respectively. 

If,  however,  the  ball  have  to  return  to  the  point  of  projection, 
we  must  have  DE  =  0,  or  the  direction  of  projection  parallel  to 
either  diagonal;  there  being  thus  two  directions  and  their  op- 
positcs,  or  four  directions  in  all,  which  will  bring  the  ball  back 
to  its  point  of  projection.  Through  this  point  it  will  pass  after 
making  each  round  of  the  table. 


DYNAMICS.  45 

1851. 

1.  A  body  of  given  elasticity  e  is  projected  along  a  hori- 
zontal plane  from  the  middle  point  of  one  of  the  sides  of  an 
isosceles  right-angled  triangle,  so  as,  after  reflexion  at  the 
hypothenuse  and  remaining  side,  to  return  to  the  same  point ; 
shew  that  the  cotangents  of  the  angles  of  reflexion  are  e  +  1 
and  e  +  2,  respectively. 

Let  ABC  (fig.  42)  be  the  triangle,  right-angled  at  -4  ;  i>  the 
middle  point  of  AB  the  point  of  projection :  AD  —  DB  =  a. 
Let  e  be  the  modulus  of  elasticity.  Draw  DEF  perpendicular 
to  BC^  making  EF  =  e.ED:  draw  FGH  perpendicular  to  ylC, 
making  GH  =  e.GF:  di'aw  HD,  LF,  KB:  DEL  will  be  the 
path  of  the  body. 

The  angle  of  reflexion  at  -ff"  =  90°  —  LEG  =  6  suppose, 

L=       LEG      =</)  

Now  BD  =  «,  .-.  BE=  ^,  =-.  EL 

'  2*        e         ' 

.-.  /6'=2a.2i-  [l+e)  BE, 


=  2a.2i-(l  +  .)-j, 

.-.  AG  =  2a-  CG=  (l+e)K 
and  FI  =  ae,   IG=  CG  =  {S-  e)  ^a, 
.-.  HG  =  e.FG  =  e  (3  +  e)  ^a, 

,._GH_  GH+AD_  l  +  ^e(3  +  g)a 

.-.    COt(^_  ^^-  ^^  -        ^(i_^g)„ 

^(2  +  e)(l-he) 
l+e         ' 

=  2+e  (1). 


46  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Again,  cot  0  =  tan  LKC  =  tau  (45°  +  KFI) , 

1  +  tarn  KFI 


1  -  ts^u KFI' 

1  +  e  t&n  LEG 

~  1-e  tsmLEG ' 

e 

'-.le 

2  +  2e 

~       2      ' 

=  1  +  e 

1) 

and 

(2) 

give 

the 

results  required. 

(2), 


2.  If  a  heavy  body  be  projected  in  a  direction  inclined  to 
the  horizon,  shew  that  the  time  of  moving  between  two  points 
at  the  extremities  of  a  focal  chord  of  the  parabolic  path  is  pro- 
portional to  the  product  of  the  velocities  of  the  body  at  the  two 
points. 

Let  S  (fig.  43)  be  the  focus  of  the  parabola,  PSj)  the  focal 
chord,  mKM  the  directrix,  KAS  the  axis ;  draw  PiV,  pn  per- 
pendicular to  the  axis,  PM^  jim  to  the  directrix.  Then,  since 
the  body  is  acted  on  by  no  horizontal  force,  its  horizontal 
velocity  will  be  constant,  and  therefore  the  time  of  moving 
from  P  to  10  will  be  proportional  to  PN+pn.  Also  (velocity)'''  at 
P  =2g.PM^  (velocity)'"'  at^  =  1g.im\  consequently  the  problem 
is  solved  if  we  can  shew  that  {PN -Vpi)^  \  PM.pm  is  a  constant 
ratio. 

Now  PM  =  8K+  SP  cobPSN  =  2AS  +  PM cosPSN, 

.-.  PM  (1  -  co&PSN)  =  2AS. 

Similarly,  pm  (1  +  cos jpSn)  =  pn  (1  +  cosPSN)  =  2A8j 

.-.   PM.pm  Bm'PSN  =  ^AS' (1). 


1851.]  DYNAMICS.  47 

smPSN 


Also  PN  =  SF  smFSN  =  PM smPSN  =  2AS 


I- cos  PSN 
from  above. 


Similarly,  pn  =  2Ab nci\T==  2 AS- ttcptt'-) 

•^ '  -^  1  +  cosPSN  1  +  cosP/SiV 

TiAT  .  A  o     sinP/SiV  ,   .  ^        1 

.-.    PX+mi  =  AAS  , 2novr=  ^^'S^    .     noAr; 

^  1  -  cos" PSN  sm  PSN  ^ 

.-.  (P.V+^n)^  sm-'P/S'^Vzz^  16^^^  (2). 

From  (1)  and  (2), 

{PN-\-p7iY  :  PM.pn  ::  4  :  1,  a  constant  ratio.     Q. e. d. 


48 


NEWTON. 

1849. 

1.  The  circle  described  through  two  points  of  an  equiangular 
spiral  and  the  point  of  intersection  of  the  tangents  at  those 
points  will  pass  through  the  pole.  Prove  this,  and  apply  tlie 
proposition  to  shew  that  the  curvature  at  any  point  of  an  equi- 
angular spiral  varies  inversely  as  the  distance  of  the  point  from 
the  pole. 

In  the  equiangular  spiral  the  tangent  is  inclined  at  a  constant 

angle  to  the  radius  vector ;  hence  in  (fig.  44)  if  P^  T,  P^  T  be  the 

tangents  at    the    points  P^P^^    S  the  pole  of  the  equiangular 

spiral  P^P^, 

SP^T+  8PJ=TT, 

and  a  circle  can  be  described  about  the  quadrilateral  SP^TP^^ 
or  a  circle  passing  through  P„  T,  P^,  will  also  pass  the  pole  S. 

Suppose  P^P^  to  be  indefinitely  near  to  each  other,  then  P^  T 
ultimately  becomes  equal  to  i^^ T,  since  the  triangles  SP^T^  STP^ 
ultimately  become  similar  and  equal.  Produce  SP^  to  meet  P^,  T 
in  B,  and  draw  P^R'  perpendicular  P^T;  then,  ultimately, 

P^E'  =  P^EsmSP^T. 
Now  diameter  of  curvature  at  P^ 

=  hmit  ^  =  ^^p-^  hmit  -^^  , 

=  g.^^py  limit     ^'^       by  the  above  property, 
2P,S 


sin  SP^T' 

or  the  curvatm'e  at  P,  varies  inversely  as  /SP,,  since  SP^  T  is 
a  constant  angle. 


NCWToX.  49 

1850. 

1.  If  any  number  of  particles  be  movhig  in  an  tllipse  about 

a  force  in  the  centre,  and  the  force  suddenly  cease  to  act,  shew 

/  1  \'^  .         . 

that  after  the  lapse  of  I  r—  ]     part  of  the  period  of  a  complete 

revolution,  all  the  particles  will  be  in  a  similar,  concentric,  and 
similarly  situated  ellipse. 

The  velocity  at  any  point  P  (fig.  45)  of  the  orbit  =  fM^CD^ 

2ir  /I  \-^ 

and  the  time  of  revolution  —  ;  therefore  after  the  [  —  )    part  of 

a  revolution,  each  particle  wiU  have  described  a  space  PP'  equal 
and  parallel  to  CD.  If  therefore  we  complete  the  parallelogram 
PCD,  P'  will  be  its  angular  point. 

Join  CP'  meeting  the  ellipse  in  Q,  and  PD  in  V.  Then,  by 
a  known  property  of  the  ellipse, 

CV.CF  =  CQ\ 
and    CP'  =  2Cr; 
.-.    CF''  =  2GQ% 
and    CP'  =  2iCQ; 
therefore   all    the    particles    are    in    a    concentric,    similar,    and 
similarly  situated  ellipse. 

2.  Two  perfectly  clastic  balls  are  moving  in  concentric 
circvdar  tubes  in  opposite  directions  and  with  velocities  pro- 
portional to  the  radii :  at  an  instant  when  they  are  in  the  same 
diameter  and  on  opposite  sides  of  the  centre,  the  tubes  are 
removed  and  the  balls  move  in  ellipses  mider  the  action  of 
a  force  of  attraction  in  the  common  centre  of  the  circles  vaiying 
inversely  as  the  square  of  the  distance.  After  one  has  per- 
formed in  its  orbit  a  complete  revolution  and  the  other  a 
revolution  and  a  half,  a  direct  collision  takes  place  between 
the  balls  and  they  interchange  orbits.  Find  the  relation  between 
the  radii  of  the  circles  and  between  the  masses  of  the  balls. 

Let  r^,  r,^  be  the  radii  of  the  circles.  Then  the  greatest  and 
least  distances  in  the  two  orbits  will  be  ?•,,  r^  in  the  first  and 
r„,  r^  in  the  second,  where  i\  has  to  be  determined. 

E 


50  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

Now  let  hji.^  be  the  values  of  h  in  the  two  orbits,  therefore 

Aj      vel.  in  circle  rad.  i\  x  r^ 
\  ~  r^^r^ ' 

=  ^ ('). 

a 

since  the  velocities  are  proportional  to  the  radii. 

h  '■'      latus-rcctum  in  first  orbit 
^^'^  t'=  second...' 

2 

=  ^'i  +  ^'^  =  !^ .  ''^  +  '"3 

'  •/  a  _  3        1^2 

^2    +    ^ 

.     !\!  =  !\    ^'2  +  ^3  ^2) 

••      ^4  ,.     •    ;.      _|_    ,. ^     ''• 

'2  '31'  2 

Also  the  periodic  time  in  the  first  orbit  =  f  that  in  the  second,  or 

-m^(!f '«)^ 


therefore,  from  (2)  and  (3), 


r;  _  /3M  r. 


r*       \2J    r. 


r,  _  /3M  r. 


and,  from  (3), 


-©*-(i)H-(i)n-(i)-"' 


1851.]  XEVVTOX.  ')l 


r. 


the  equation  tor  finding  —  .     This  equation  has  only  one  positive 

root,  and  that  less  than  1,  as  it  ought  to  be,  n  suppose. 

To  find  the  relation  between  the  masses  m^  and  nr,  of  the 
balls  in  the  greater  and  less  orbit  respeetively.  Let  i\  and  ?'.^ 
be  tlieii'  velocities  before  impact ;  their  velocities  after  impact 
will  be  I'.,  and  i\  respectively,  m^  and  m,,  both  moving  after 
impact  in  the  same  direction  as  m^  the  greater  did  before 
impact.  Hence,  since  the  elasticity  is  perfect,  momentum  lost 
by  7»j  =  the  whole  momentum  lost  and  gained  by  vn^,  or 

and  since  the  balls  are  at  the  same  distance  from  the  centre  of 
force,  and  moving  in  opposite  directions, 

^  =  ^  =  i'  by  (1) 

2  2  2 

1 


V 

-^  +  1 

-1-1 

^2 

1    +    )l' 

the  required  relation  between  the  masses. 

1851. 

1.  If  a  body  describe  an  ellipse  round  a  centre  of  force  in 
the  focus,  shew  that  the  sum  of  the  reciprocals  of  the  squares  of 
the  velocities  at  the  extremities  of  any  chord  passing  through 
the  other  focus  is  constant. 

Let  PHp  (fig.  46)  be  the  chord  thi'ough  II.  Draw  the  pi-r- 
pendiculars  SY^  St/,  HZ,  Hz,  to  the  tangents  at  those  points: 
join  SP,  Sj). 

e2 


52  SoLlTlONb   OF   SliNATE-lluUSl-:    PROISLEMS.  [1851. 

Then,  by  a  known  property, 

1  14 

^^^  +  T7-  =  T  [L  the  latus-rectum), 
JJ.F      lip      L 

2AC      ,      2AC      ,_B.^C 
SP      Sp       SAC      . 
SY      Sy      SAC      ^ 

or  •.•  SY.HZ^  SyJIz  =  BC'\ 

SY^      Sjf_^SAM 
BC  "^  BC  ~     L  ' 

or    SY^  +  Sy^  is  constant, 

and  the  velocities  at  -P,^?,  are  inversely  proportional  to  8Y^  Sy, 
therefore  sum  of  the  squares  of  the  reciprocals  of  the  velocities 
at  Pp  are  constant. 

Cor.  It  may  also  be  shewn  that  the  sum  of  the  squares  of 
the  velocities  at  the  extremities  of  any  chord  passing  thi'ough 
the  centre  of  force  is  constant. 

For  we  have  shewn  that 

SY      Sy  ,     , 

^^-^  +  -~  =  constant, 
HZ      Hz  ' 

SY.HZ      Sy.Hz  ^    ^ 

•••      ^^2     +    jj^^     =  constant, 

or   BC'^  [iTt''  "*"  77^ )  ~  constant, 

'*'  TiZ'  "*"  'Hz^  ^  constant, 

or,  taking  H  as  the  centre  of  force,  the  siun  of  the  squares  of  the 
velocities  at  the  extremities  of  any  chord  passing  through  the 
centre  of  force  is  constant.* 

*  TTiis  corollary  was  set  as  a  problem  in  1848. 


I 


(     53     ) 


HYDROSTATICS. 

1848. 

1.  An  inverted  vessel  formed  of  a  substance  which  is  heavier 
than  water  contains  enough  of  air  to  make  it  float :  prove  that 
if  it  be  pushed  down  througli  a  certain  space,  it  will  be  in  a 
position  of  unstable  equilibrium ;  and  deteraiine  the  space  in 
question. 

When  the  vessel  is  floating  partly  immersed,  the  weight  of 
the  water  displaced  is  equal  to  the  weight  of  the  vessel  and  of 
the  air  it  contains.  If  the  vessel  be  now  pushed  down,  the 
water  displaced,  and  therefore  the  upward  pressure  on  the  vessel, 
will  be  increased  till  the  vessel  is  wholly  immersed ;  as  the 
vessel  is  now  pushed  down  further  the  water  displaced  becomes 
less  on  account  of  the  compression  of  the  air  in  the  vessel,  till 
it  comes  into  such  a  position  that  the  weight  of  the  water  dis- 
placed is  only  equal  to  the  weight  of  the  vessel  and  the  air  it 
contains.  This  will  be  a  position  of  equilibrium ;  and  the  equi- 
libi-ium  will  be  mistable,  for  accordingly  as  it  is  a  little  above 
or  a  little  below  tliis  position,  the  weight  of  the  water  displaced 
will  be  greater  or  less  than  that  of  the  vessel  and  the  air  it 
contains. 

This  explanation  applies  to  a  vessel  of  a  cylindrical  form ; 
if,  however,  it  is  smaller  at  the  top  than  the  bottom  it  may  come 
into  the  position  of  unstable  equilibrium  before  it  is  wholly 
unmersed.  To  find  how  far  the  vessel  must  be  displaced  so  as 
to  come  into  this  position. 

Let  W  be  the  weight  of  the  vessel,  V  its  volume ;  a  and  j- 
the  altitude  of  the  column  of  air  in  the  vessel  in  the  positions 


54  SOIA'TIONS   (IF   SENATE-HOUSE    PROBLEMS.  [1848. 

of  Stable  and  luistablc  equilibrium,  V'V"  its  volumes  in  those 
positions,  1/  the  depth  of  its  lower  surface  below  the  surface  of 
the  water  in  the  latter  position:  a  the  density  of  the  water, 
p,  p  those  of  tlie  air  in  the  two  positions,  jjy  p  its  pressures  in  the 
same  positions. 

Then,  equating  the  weight  of  the  fluid  displaced  and  of  the 
vessel  and  air  contamed, 

^TV=  W+<jV"p' (1). 

Also  equating  the  upward  pressure  of  the  water  and  downward 
pressm-e  of  the  air  at  their  common  surface, 

!/(^!/  =P'  (2)- 

Also,  since  pressm'e  and  density  vary  inversely  as  volume, 

^  =  ^  =  {; (3). 

Again,  when  the  form  of  the  vessel  is  known,  V  and  V"  will  be 
known  in  tenns  of  a  and  .r.  Hence  equations  (1),  (2),  and  (3), 
will  be  sufficient  for  the  determination  of  x  and  ?/,  as  well  as  the 
other  unknown  quantities,  viz.  ^j',  p',  and  V".  Hence  y  —  x^  or 
the  depth  of  upper  surface  of  the  air  below  that  of  the  Avater,  is 
knoAvn,  and  added  to  the  difference  of  the  heights  of  the  same 
two  surfaces  in  the  original  position  of  equilibrium,  gives  the 
space  through  which  the  vessel  must  be  depressed. 

2.  A  uniform  piston,  terminated  by  a  plane  of  area  A^  per- 
pendicular to  its  side,  is  inserted  into  an  orifice  in  a  vessel 
containing  fluid ;  prove  that  the  work  done  in  gently  pushing 
in  the  piston  through  a  small  space  s  is  ultimately  equal  to  the 
work  done  in  lifting  a  portion  of  the  fluid  of  volume  As  through 
a  height  equal  to  the  depth  of  the  centre  of  gravity  of  the  plane 
below  the  surface  of  the  fluid. 

If  the  space  s  be  indefinitely  small,  the  pressure  on  each 
element  of  the  piston  will  l)o  unaltered  by  the  change  of  the 
pistoirs^  position. 


1849.]  HYDROSTATICS.  55 

Hence,  i(  s  be  indefinitely  small, 

work  done  =  product  of  whole  pressure  on  area  A  x  s^  the  space 
through  which  it  is  moved  perpendicular  to  itself, 

=  pressure  at  depth  z  of  the  centre  of  gravity  of 
A  X  area  A  x  Sy 

=  gpzA^y 

=  ffpAs.z, 

=  weight  of  volume  As  of  the  fluid  x  2, 

=  work  done  in  raising  the  volume  As  through  the 
space  z. 

3.  Two  equal  slender  rods  ABj  AC^  moveable  about  a  hinge 
at  Ay  and  connected  by  a  string  BC^  rest  with  the  angle  A 
immersed  m  a  given  fluid ;  determine  the  tension  of  the  string 
BC. 

Let  T  =  tension  of  the  string, 
w  =  weight  of  each  rod, 
2a  =  its  length, 

2l  =  the  length  of  the  part  immersed, 
a  =  its  inclination  to  the  horizon. 
Then  the  rod  is  kept  at  rest  by  its  weight,  the  tension  of  the 
string,  the  action  at  the  hinge  and  the  fluid  pressures  which 
have  for  resultant  a  vertical  upward  pressure   w  acting  at  a 
distance  I  from  A. 

Hence,  taking  moment  about  A, 

10 .a  cos  a  —  w.l  cos  a  —  T.2a  sin  a  =  0, 


1  =  -— —  coicnAv 
2a 


is  the  required  tension. 


1849. 

A  body  floats  in  a  mixture  of  two  given  fluids  with  a  volume 
A  immersed;  one  half  of  the  mixture  being  removed,  and  its 
place  supplied  by  an  equal  quantity  of  the  lighter  fluid,  the 
same  bodv  floats  Avith  a  volume  A  +  B  immersed.     Determine 


56  SOLUTIONS   OK   SENATE-HOUSE    PROBLEMS.  [1850. 

the  ratio  ot"  the  quantities  of  fluid  in  the  original  mixture,  sup- 
posing the  volume  of  the  mixture  to  be  equal  to  the  sum  of  the 
volumes  of  the  eomponcnt  fluids. 

Explain  the  result  when  the  densities  of  the  fluids  are  as 
^  +  ^  to  ^  -  ^. 

Let  F,  V  be  the  original  volumes  of  the  fluids ;  cr,  a'  their 
specific  gravities.  Their  volumes  in  the  second  mixture  will  be 
^V  and  ^V  +  ^{V  +  V)  or  V  +  ^T';  the  specific  gravities  of 
the  mixtures  will  be 

hence,  if  W  be  the  weight  of  the  body, 

PWFV 

n   -A      ^.^^.     , 

also    =  [A  +  H]  p  _^  y ) 

.-.   A[Vct+V'<t')  =  {A+B)  [iFo-  +  (F  +  iF)(7'}, 

.-.    {AcT-^{A  +  B)a-^{A+B)a'}  V^Ba'V, 

V  2B(r' 

or 


F  '  [A-B]  (7-{A  +  B)a" 

the  required  ratio. 

If  tlie  densities,  and  tlierefore  the  specific  gravities,  are  as 
A  +  B  io  A  —  B,  F'  =  0,  shewing  that  the  fluid  cannot  be  a 
mixture  of  fluids  of  diiferent  specific  gravities ;  in  fact,  the  con- 
ditions of  the  problem  then  become  impossible. 

1850. 

1.  A  conical  vessel  containing  a  given  quantity  of  fluid  has  Its 
axis  vertical,  and  another  cone  with  the  same  vertical  angle  is 
placed  to  float  in  the  fluid  with  its  vertex  downwards ;  find  how 
much  the  fluid  will  rise  in  consequence. 

Let  h  be  the  depth  of  the  original  cone  of  fluid,  k  the  depth 
to  which  the  vertex  of  the  floating  cone  will  sink  ;  A-  is  known 
from  the  specific  gravities  of  the  fluid  and  floating  cone,     z  the 


1850.]  HYDROSTATICS.  57 

height  thi'ough  which  the  fluid  will  rise.     Then 

volume  of  the  cone  height  /<!  +  s       {h-\-  zf 
h       "      1i'      ' 


,     volume  of  the  cone  height  k      k 

and 7  =  T?  • 

h      h^ 

Therefore,  subtracting, 

volume  of  the  fluid  after  the  cone  Is  put  Into  It  _  (A  +  zf  —  k^ 
its  original  volume  K^  ' 

=  1, 
since  the  quantity  of  fluid  is  unaltered ;  therefore 

{h  +  zf  -k'  =  F 

and   z  =  {h'  +  k'Y  -  h 
is  the  required  space. 

2.  A  hollow  cylinder  containing  air  Is  fitted  with  an  air- 
tight piston  which,  when  the  cylinder  is  placed  vertically,  is  at 
a  given  height  above  the  base ;  the  cylinder  being  now  inverted 
and  placed  vertically  in  a  fluid,  sinks  partly  below  the  smiace ; 
find  the  position  of  equilibrium. 

Let  J)  be  the  pressure  of  the  air  in  the  cylinder  before  the 
cylinder  Is  inverted,  and  which  Is  knoAvn  from  the  given  height 
(/«)  of  the  piston  above  the  base :  p  is  the  pressure  due  to  the 
weight  of  the  piston  and  atmosphere,  11  the  atmospheric  pres- 
sure, to  the  weight  of  the  cylinder  and  piston,  z  the  depth 
below  the  sm*face  of  the  fluid  of  the  piston  In  the  position  of 
equilibrium,  i/  the  distance  of  the  piston  from  the  base  of  the 
cylinder,  p,  p  the  densities  of  the  fluid  and  uncompressed  air ; 
then,  for  the  equilibrium  of  the  piston, 

fluid  pressure  fi'ora  beneath  =  pressure  due  to  the  weight  of  the 
piston  +  the  pressure  of  the  air  in  the  inverted  cylinder, 

ov  c/pz  +  U  =2^-U  +  -i> (1). 


58  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Also  ic  +  weight  of  the  air  iii  the  eyliiider 

=  weight  of  the  fluid  displaced, 

.-.    w+gp'V  =  fjpj^,  V (2), 

where  Via  the  volume  and  //  the  height  of  the  cylinder. 
From  equation  (2)  z  is  known,  and  thence  y  from  (1). 

1851. 

1.  A  hollow  cone  floats  m  a  fluid  with  its  vertex  upwards 
and  axis  vertical ;  determine  the  density  of  the  air  in  the  hollow 
cone. 

Let  2^  be  the  pressiu'e  of  the  air  in  the  cone,  IT  that  of  the 
atmosphere,  tv  the  weight  of  the  cone,  h  its  height,  z  the  height 
of  the  cone  of  compressed  air,  y  the  depth  of  its  base  below  the 
surface  of  the  fluid,  /?,  p'  the  densities  of  the  fluid  and  uncom- 
pressed air. 

Then,  equating  the  pressures  at  the  common  smface  of  the 
air  and  fluid, 

gpy  +'n  =  pressure  of  the  compressed  air, 

=i>  =  -n (1). 

Also  10  +  weight  of  the  air  in  the  cylinder 

=  weight  of  the  fluid  displaced, 

,TT           z^  —  (z  —  vY  _-  ,  , 

or   to  +  gpV=gp Jf-^  ^  (2), 

where  V  is  the  volmne  of  the  cone. 

From  equations  (1)  and  (2)  z  and  y  are  known,  and  thence  j-j, 
and  the  required  density. 


(  '^9  ) 


OPTICS. 

1848. 

1.  If  Q^  q  (fig.  47)  be  two  points  in  the  radius  of  a  spherical 
reflecting  surface  whose  centre  is  E^  such  that  EQ  :  Eq  ::  sine 
of  the  angle  of  incidence  :  sine  of  the  angle  of  refraction,  de- 
termine geometrically  the  position  of  the  point  P,  so  that  a  ray- 
proceeding  from  Q  and  incident  upon  the  surface  at  P  may 
after  refraction  proceed  from  q. 

Bisect  Qq  in  m,  and  thi'ough  m  draw  mF  perpendicular  to 
EQ  meeting  the  circle  in  P;  Pwill  be  the  point  required.  For 
if  we  join  FE^  Fq^  FQ,  we  have 

sin^P^  :  sin  FQE::  QE :  FE, 
and   sm^P^  :  sinP<^^  ::  qE  :  FE. 
Now      sin  FQE  =  sin  FqE,     •.  •  z  FQE  =  LFqQ, 
.-.  smEFQ  :  sinJ;P^  ::  EQ  :  Eq  ::  fM  :  I 
by  the  question,  therefore  the  ray  QF  after  refraction  at  P  will 
proceed  as  if  from  q. 

2.  If  a  ray  of  light,  after  being  reflected  any  number  of 
times  in  one  plane,  at  any  nimiber  of  plane  sm'faces,  retmii  on 
its  fonner  course,  prove  that  the  same  will  be  true  of  any  ray 
parallel  to  the  foi'raer  which  is  reflected  at  the  same  surfaces 
in  the  same  order,  provided  the  number  of  reflections  be  even. 

Let  FQR8  (fig.  48)  be  the  course  of  any  ray  which  starting 
from  P,  after  reflection  at  Q^  R  and  8^  amvcs  again  at  P, 
and  is  there  reflected  in  the  direction  PQ  of  the  original  pro- 
pagation. Let  F'Q'JR'S'  be  the  course  of  another  ray  starting 
from  P'  in  a  dii'ection  F' Q'  parallel  to  FQ',  we  have  to  shew 
that  after  reflection  at  5",  this  ray  will  proceed  to  P,  and  there 
be  reflected  in  the  direction  P'  Q'.     Join  S'F'. 

Then  since  the  angle  Q'F'A  =  QPA,  and  QPA=SFF,  there- 
fore the  triangle  PpP'  is  isosceles,  and  the  ptTpciidiiiilar  frmn  /' 


60  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS,  [1849. 

on  F  Q  =  that  from  P'  on  PS.  Similarly  the  perpendicular 
from  Q  on  QR  =  that  from  Q  on  Q'P'  =  that  from  P  on  P  Q 
since  PQ  is  parallel  to  P'  Q ;  therefore  the  pei'pendicular  from 
Q  on  QR  =  that  from  P  on  PS.  By  similar  reasoning  it  may 
be  shewn  that  the  pei'pendicular  from  S'  on  PS  =  that  from  R' 
on  QR  =  that  from  Q'  on  ^i?  since  Q'R'  is  parallel  to  ^^  =  that 
from  P  on  PS.  Hence  PS'  is  parallel  to  PS'^  therefore  S'P'  is 
the  direction  in  which  R' S'  will  be  reflected  from  S\  and  P  Q' 
is  that  in  which  S'P  will  be  reflected  from  P'. 

The  same  proof  may  be  extended  to  any  even  nmnber  of 
reflections.  If  the  number  of  reflections  were  not  even  we  might 
still  shew  that  P',  S'  were  equidistant  from  PS^  but  they  would 
be  on  opposite  sides  of  it,  as  P',  R'  are  of  PP,  and  the  pro- 
positions would  not  be  true  in  that  case. 

1849. 

If  the  angle  of  a  hollow  cone,  polished  internally,  be  any 
submultiple  of  180°,  a  cylindrical  pencil  of  rays  incident  parallel 
to  the  axis  will,  after  a  certain  number  of  reflections,  be  a 
cylindrical  pencil  parallel  to  the  axis,  and  of  the  same  diameter 
as  the  incident  pencil. 

Let  fig.  49  represent  a  section  of  the  cone  and  the  light  by  a 

plane  through  the  axis  CD  of  the  cone,  and  let  tn^m_^ be  the 

successive  points  when  the  ray  PQ^Q^Q^...  cuts  the  axis  CD. 

180° 
(1).  Let  the  angle  ACB  be  an  even  submultiple  of  180°=— — 

90°  ^'' 

suppose,  or  —  . 

Now  the  angle    Q^m^D  =  ACD  +  CQ^m^  =  ACD  +  A  Q^P, 
=  2ACD  =  ACB, 
and  the  angle   Q^^n.^D  =  BCD  +  CQ,^in^  =  BCD  +  Q^Q^B, 
=  BCD  +  BCD  +  Q.m^D, 
=  2ACB- 
similarly   Q^m^D  =  3 A  CB, 


~  ) 

and   Q„<»,D  =  uACB  =  90°, 


1850,]  OPTICS.  61 

or  after  the  »"'  reflection  the  ray  will  be  pci*pendicular  to  the 
axis  CD^  and  will  proceed  in  a  path  exactly  similar  to  that 
already  described,  finally  emerging  in  a  direction  parallel  to 
QyP^  and  at  the  same  distance  as  Q^P  from  the  axis  CD^  but 
on  the  opposite  side  of  it. 

180° 

(2).  Let  the  angle  A  GB  be  an  odd  submultiple  of  180°= 

^  >  ^  ^  2n+l 

suppose. 

Now  the  angle   Q.Q^B  =  BCD  +  Q,m,D^ 

=  IACB  -\-  ACBhj  the  above, 

and  the  angle    Q.^Q^A  =  A  CD  +  Q.^mJ)^ 
=  \ACB  +  2ACB, 
=  IACB; 
similarly    Q.^Q^B  =  IACB, 

and   Q,^Q,^,^A  =  ^-^ACB  =  dO'', 

or  after  w  reflection  the  ray  will  be  perpendicular  to  the  side 
CA  or  CBj  at  which  it  has  next  to  be  reflected,  and  will  there- 
fore after  that  reflection  return  by  the  same  path  as  it  came  by, 
and  will  emerge  in  the  direction  Q^P. 

Hence,  whether  ACB  be  an  even  or  odd  submultiple  of  180°, 
the  emergent  rays  will  form  a  cylinder  equal  in  diameter  to  the 
cylinder  of  incident  rays,  and  having  its  axis  coincident  with  the 
axis  of  that  cylinder,  if  the  angle  A  CB  be  an  odd  submultiple 
of  180°;  or  if  the  angle  ACB  be  an  even  submultiple  of  180", 
the  axes  of  the  emergent  and  incident  pencils  will  lie  in  the 
same  plane  with  the  axis  of  the  cone  at  equal  distances  on 
opposite  sides  of  it. 

1850. 

1.  If  a  luminous  point  be  seen  after  reflection  at  a  plane 
min'or  by  an  eye  in  a  given  position,  there  is  a  certain  space 
within  which  the  image  of  the  point  can  never  be  situated,  how- 
ever the  position  of  the  plane  mirror  be  changed ;  find  this 
space. 


62  SOLUTIONS   OF    SKXATK-Ilorsi:    I'KOBl.EMS.  [iSol. 

It  is  easily  seen  that  the  distance  from  the  eye  of  the  image 
foiTned  by  the  mirror  equals  the  actual  length  of  the  ray  by 
which  the  point  is  seen.  This  can  never  be  less  than  the  direct 
distance  of  the  point  from  the  eye ;  hence  the  image  can  never 
be  situated  within  the  sphere  which  has  the  dii'ect  distance 
between  the  point  and  the  eye  for  radius. 

2.  If  a  be  the  angle  which  every  diameter  of  a  circular 
disc  subtends  at  a  luminous  point,  shew  that  the  ratio  of  the 
light  Avhich  falls  on  the  disc  to  the  whole  light  emitted  is  as 
sin'^^a  :  1. 

About  the  Imninous  point  as  centre  describe  a  sphere  with 
radius  unity :  also  with  the  luminous  point  for  vertex  and  the 
circular  disc  as  base  describe  a  right  cone.  Then  the  light 
received  on  the  circular  disc  :  whole  light  emitted  ::  the  portion 
of  the  surface  of  the  sphere  Intercepted  by  the  cone  :  whole 
surface  of  the  sphere. 

Now  by  a  known  property  of  the  sphere,  the  surface  of  any 
portion  of  the  sphere  cut  off  by  any  plane  is  proportional  to  the 
difference  of  the  radius  of  the  sphere  and  the  distance  of  the 
cutting  plane  fi'om  the  centre.  Hence  the  surface  intercepted 
by  the  above  cone  :  whole  surface  of  the  sphere  ::  1  —  cos|a 
:  2  ::  sin^'ja  :  1,  which  is  therefore  the  ratio  of  the  light  received 
on  the  circular  disc  to  the  whole  light  emitted. 

1851. 

A  sphere  composed  of  two  hemispheres  of  different  refrac- 
tive powers  is  placed  in  the  path  of  a  pencil  of  light  in  such 
a  maimer  that  the  axis  of  the  pencil  is  perpendicular  to  the 
plane  of  jimctlon  and  passes  through  the  centre :  determine  the' 
geometrical  focus  of  the  refracted  pencil. 

Let  r  be  the  radius  of  the  sphere,  u  the  distance  of  the  focus 
of  incident  rays  from  the  centre;  v^v^v^  the  distances  of  the 
geometrical  foci  after  the  successive  refractions,  positive  lines 
being  measured  m  the  direction  opposite  to  that  of  the  incident 
light ;  fi^fx,,  the  refractive  indices  of  the  two  hemispheres. 


1851.]  OPTICS.  63 

Then  i  =  -  ^Vzi  +  ^. (1), 

1       /i,    1 

-  =^  - (2  , 

(3)  +  (2)xl  +  (l)x^,, 

i=-^.i^>,-i)+^,>,-i)}i+^:i, 

which  gives  \\  the  distance  from  the  centre  of  the  sphere  of  the 
geometrical  focus  after  refraction. 


(  <>4  ) 


ASTRONOMY. 

1849. 

1.  Tlierc  arc  two  walls  of  equal  known  height  at  right 
angles  to  each  other,  and  running  in  known  directions ;  shew 
how  to  find  the  sun's  altitude  and  azimuth  by  observing  the 
breadth  of  the  shadows  of  the  two  walls  at  any  given  time. 
And  prove  that  the  sum  of  the  squares  of  the  breadths  of  the 
shadows  will  be  the  same  whatever  be  the  direction  of  the  Avails. 

Let  a,  h  be  the  observed  breadths  of  the  shadows,  h  the 
known  height  of  the  walls ;  6  the  angle  between  the  base  of 
the  wall,  the  breadth  of  whose  shadow  is  a,  and  the  line  joining 
the  shadow  of  the  top  of  the  line  of  intersection  of  the  walls 
with  the  bottom  of  that  line,  ^  the  sun's  altitude.     Then 

(y  =  tan     7- ,    and    ©  =  tan     j-^ ^,-^, , 

i  '  ^  (a^  +  ly 

are  known.     Let  a  be  the  angle  between  the  wall  whose  breadth 

is  a  and  the  plane  of  the  meridian ;    then   a  --  ^  is  the  angle 

between  the  plane  of  the  meridian  and  the  vertical  plane  through 

the  sun,  or  the  sun's  azimuth.     Hence  both  the  altitude  and 

azimuth  are  known. 

Also  d^  +h^  =  the  square  of  the  length  of  the  shadow  of  the 

Ime  of  intersection  of  the  walls ;  and  the  height  of  this  line  is 

the  same  whatever  be  the  direction  of  the  walls,  or  a''  +  //  is 

independent  of  that  direction. 

2.  If  the  same  two  stars  rise  together  at  two  places,  the 
places  will  have  the  same  latitude.  And  if  they  rise  together 
at  one  place  and  set  together  at  the  other,  the  places  will  have 
equal  latitudes,  but  one  north  and  the  other  south. 

From  the  bisection  of  SS'  the  great  circle  passing  through 
the  two  stars  S^  S'  draw  a  quadrant  of  a  great  circle  perpen- 
dicular  to    SS'    towards   the   north    pole    terminating    in    the 


1850.]  ASTRONOMY.  (55 

point  r,  and  another  towards  the  south  pole  terminating  in  T. 
First,  suppose  the  great  circle  containing  these  quadrants  to 
have  its  point  which  is  nearest  to  P  the  north  pole,  to  tlic  left 
of  P (drawing  the  stars  on  the  convex  part  of  the  sphere).  Then, 
in  order  that  the  stars  /S',  S'  may  rise  together  at  any  place, 
its  zenith  must  at  some  time  in  the  24  hours  come  to  T\  its 
co-latitude  therefore  must  be  TP.  Hence  if  S^  8'  rise  together 
at  two  places,  the  co-latitude  of  each  must  be  2!P;  hence  the 
latitudes  of  the  places  are  the  same.  If  the  point  of  the  circle 
nearest  to  P  lie  to  the  right  of  P,  the  zenith  of  the  place  must 
pass  through  T'  in  their  daily  path  and  therefore  have  the  same 
latitude,  viz.  90'  —  T'P\  S.,  P'  being  the  south  pole. 

If  the  same  stars  rise  together  at  one  place  and  set  together 
at  another,  the  zenith  of  one  must  pass  tlii'ough  T  and  that 
of  the  other  through  T  m  their  daily  paths;  hence  they  will 
still  have  equal  latitudes,  but  one  will  be  north  and  the  other 
south. 

1850. 

1.  Prove  that  all  stars  which  rise  at  the  same  Instant  at 
a  place  within  certain  limits  of  latitude,  will,  after  a  certain 
interval,  lie  in  a  vertical  great  circle ;  and  detenuine  those 
limits. 

This  Avill  happen  when  the  zenith  of  the  place  comes  to  that 
great  circle  of  the  heavens  which  at  the  time  of  the  stars'  rising 
was  the  horizon  of  the  place.  Hence  it  can  only  happen  for 
those  places  for  which  the  altitude  of  the  pole  is  less  than  the 
co-latitude :  but  the  altitude  of  the  pole  is  the  latitude,  hence 
if  /  be  the  latitude,  I  must  be  less  than  90°  —  /  or  /  less  than  A,'/. 

2.  Shew  how  to  find  the  days  of  the  year  on  which  the 
light  of  the  sun  reflected  by  a  given  window  which  has  a  south 
aspect  will  be  thrown  into  some  one  of  the  lower  windows  of  an 
opposite  range  of  buildings. 

Corresponding  to  each  window  opposite,  let  that  point  of  the 
heavens  be  detennined,  which  lies  in  the  same  plane  as  that 
window  and  the  horizontal  line  through  the  reflecting  window 
pointing  to  the  south,  and  at  the  same  angidar  distance  from  this 

r 


66  SOLUTIONS   OF   SENATE-HOUSE   PKOBLEMS.  [1851. 

line  as  the  opposite  window  in  question,  and  on  tlie  opposite 
side  of  it.  Let  the  north  polar  distance  of  this  point  or  its 
declination  be  then  observed ;  the  reflected  light  will  enter  the 
window  in  question  on  those  days  when  the  sun  has  this  de- 
clination. Suuilarly,  the  days  when  the  reflected  light  will 
enter  the  other  windows  may  be  detenniued. 

1851. 

Altitudes  of  the  same  heavenly  body  are  observed  from  the 
deck  of  a  ship  and  from  the  top  of  the  mast  the  height  of  which 
from  the  deck  is  known :  find  the  dip  of  the  horizon  and  the 
tnie  altitude. 

Let  AB  =  x,  BC=  h  (fig.  50)  be  the  height  of  the  deck 
from  the  sea,  and  of  the  mast  respectively ;  OA  =  r  the  radius 
of  the  earth.  The  difference  (a)  of  the  observed  altitudes  is  the 
angle^i^Oori>(9^. 

Now    cosCOE  = 5 ,   and  cosCOD  = , 

r  +  h  +  x^  r  +  x' 

.-.    DOE  =  cos"^ 5 cos  ^ =  a, 

r  +  li  +  X  r  +  X 

an  equation  for  the  deteinnination  of  x. 

T 

Then  the  dip  of  the  horizon  =  OBD  =  sin~^ is  known, 

and  subtracted  from  the  altitude  observed  at  B  gives  the  true 
altitude. 

From  the  above  equation  we  may  determine  x  with  sufficient 
accuracy  thus : 


r  +  X  \         {r  +  h  +  xY)        r  +  h  +  x  \         {r+x 


\2r(h  +  x)}^-       (2rx)^ 

or  - — ^ -^^ —  =  sma, 

r  r  ' 


omitting  h  and  x  in  comparison  of  r ; 


2(h  +  x)        .2         «   •       /2a;\i      2x 

-^ '  =  sm^a  -  2  sma    —     -\ , 

r  \  r  J         r  ^ 


f2x\^-      ,    .  h 

I I    =  4  sin  «  —  — 

h 


,       ,  —  9  sma coseca, 

\r  J        ^  r  ' 


or   X  =  l^  sma  — -  coseca  I 


PART  II. 


F-2 


(     69     ) 


PART    11. 


EUCLID. 

1848. 

1.  AB,  CD,  (fig.  51)  are  any  two  chords  of  a  circle  passing 
through  a  fixed  point  O,  EF  any  chord  parallel  to  AB ;  join 
CE,  DF  meeting  AB  in  the  points  G  and  H,  and  DE,  CF 
meeting  AB  in  the  points  K  and  L :  shew  that  the  rectangle 
OG.GH  =  OK.OL. 

The  triangles  OCG,  OHD  have  the  common  angle  0,  and 

^OCD  =  180°  -  EFD  =  EFH, 

=  OHD, 

since  EF  is  parallel  to  AB ;  hence  the  triangles  OCG,  OHD 
are  similar,  therefore 

OC:OG::OH:OD, 
or  OG.OH  =  OC.OD. 

Again,     L  OCD  =  OEF  =  DKO, 

since  EF  is  parallel  to  AB ;  hence  the  triangles  OLC,  ODK 
are  similar,  therefore 

OC:OL::OK:OD, 
or  OL.OK  =  OC.OD, 
.-.  OG.OH  =  OL.OK. 

2.  In  a  given  circle  inscribe  a  rectangle  equal  to  a  given 
rectilineal  figm'c. 

Let  AB  (fig.  52)  be  a  diameter  of  the  given  circle  ABC. 
Draw  a  square  which  shall  be  equal  to  the  rectilineal  figure 


70  SULI'TIONS   OF   SEXATE-HOUSE   PROBLEMS.  [1848. 

{Eu<\  II.  14,;  through  D  any  point  of  AB  draw  BE  pei*pen- 
dicular  to  AB,  a  third  projxjrtional  to  AB  and  the  side  of  the 
square.  Through  E  draw  EC  parallel  to  AB,  meeting  the  circle 
in  C ;  join  AC,  BC,  and  complete  the  parallelogram  ACBF : 
it  shall  be  the  rectangle  required. 

For  C  and  F  are  each  right  angles,  being  the  angles  in  a 
semicircle,  therefore  ACBF  is  a  rectangle.  Also  its  area  equals 
AB.DE  which  equals,  by  constniction,  the  square  which  equals 
the  given  rectilineal  figure ;  therefore  it  is  the  rectangle  required. 

3.  Through  a  given  point  A  (fig.  53)  describe  a  circle  which 
shall  touch  a  given  circle  BCD,  and  intersect  another  given 
circle  LEF  in  a  chord  passing  through  a  given  point  G. 

From  G  draw  any  line  GEF  Intersecting  the  circle  LEF  In 
the  points  E,  F  ;  join  GA,  and  in  GA  produced  if  necessaiy,  take 
the  point  H,  such  that  GA.GH  =  GE.GF.     Through  the  points 

A,  H  describe  any  circle  cutting  the  circle  BCD  In  the  points 

B,  C  ;  join  BC,  and  produce  It  to  meet  GA  In  K.  From  K  draw 
KD  a  tangent  to  the  circle  BCD.  About  the  triangle  AHD 
describe  a  circle,  it  shall  be  the  circle  required. 

And  first  It  shall  touch  the  circle  BCD :  for  since  KD 
touches  the  circle  BCD, 

KD^  =  KB.KC  =  KA.KH, 

since  one  circle  has  been  made  to  pass  through  A,  H,  C,  and  B ; 
and  therefore  KD  touches  the  cu'cle  m  question  as  well  as  the 
circle  BCD,  therefore  the  two  circles  touch.  Also  the  chord  in 
which  it  intersects  the  circle  LEF  will  pass  through  G.  For 
suppose  L  to  be  one  of  the  points  In  which  It  intersects  the  circle 
LEF ;  join  GL  and  produce  It  to  meet  the  two  circles  in  M,  M'. 
Then 

GL.GM  =  GE.GF  =  GA.GH,  by  construction; 

also  GL.GM' =  GA.GH, 

since  H,  A,  L,  M',  lie  in  the  circumference  of  the  same  circle, 
therefore  GM  =  GM'  or  the  points  M  and  M'  coincide,  and 
GLM  Is  the  chord  in  whicli  the  circles  intersect,  and  the  chord 
passes  through  G  as  required. 


1851.]  EUCLID.  71 

Hence  tlic  circle  drawn  as  above  described  fulfils  the  required 
conditions,  and  is  therefore  the  circle  sought. 

1849. 

Thi-ee  circles  are  described,  each  of  which  touches  one  side  of 
a  triangle  ABC  (fig.  54),  and  the  other  two  sides  produced.  If 
D  be  the  point  of  contact  of  the  side  BC,  E  that  of  CxV,  and  F 
that  of  AB,  shew  that  AE  =  BD,  BF  =  CE,  and  CD  =  AF. 

Let  AB,  AC  touch  the  circle  GDH  in  G,  H ;  then 
BG  =  BD,   and  CH  =  CD, 
also   AG  =  AH, 
.•.   AB  +  BD  =  AC  +  CD  =  semiperimeter  of  the  triangle, 
similarly,   BA  +  AE  =  semiperimeter  of  the  triangle  : 
.-.   BA  +  AE  =  AB  +  BD, 
and   AE  =  BD ; 
and  similarly,  BF  =  CE  and  CD  =  AF. 

1851. 

1.  Let  T  (fig.  55)  be  a  point  without  a  circle,  whose  centre 
is  C ;  from  T  draw  two  tangents  TP,  TQ ;  also  through  T  draw 
any  line  meeting  the  circle  in  V,  and  PQ  in  B,  and  draw  CS 
perpendicular  to  TV;  then  SR.ST  =  SY\ 

Join  CT,  intersecting  PQ  in  U  at  right  angles ;  draw  CP; 
it  will  be  perpendicular  to  PT. 

Since  the  triangles  CTS,  RTU  are  similar, 
CT:TS::RT:  TU, 
.-.    TS.TR  =  TC.TU, 
or   ST'^  -  ST.SR  =  CT'  -  CT.CU, 

=  ST^  +  CS^  -  CT.CU, 
.-.   ST.SR  =  CT.CU -CS^: 
but  CPT,  PUC  are  both  right  angles, 

.-.   CT.CU  =  CP''  =  CV^ 

=  SV^  +  cs% 

.-.   ST.SR  =  SV^ 


72  [SOLUTION'S   OF   SEXATE-HOUSK    PROBLEMS,  [1851. 

2.  It"  a  ciix'lc  be  described  round  the  poiut  of  intersectiou  of 
the  diameters  of  a  parallelogram  as  a  centre,  shew  that  the  sum 
of  tlie  squares  of  the  lines  drawn  from  any  point  in  its  cir- 
cmnference  to  the  four  angular  points  of  the  parallelogram  is 
constant. 

Join  P  any  point  in  the  circle  with  A,  B,  C,  D  (fig.  56)  the 
angular  points  of  the  parallelogram,  and  with  O  the  centre  of 
the  circle.  Then,  since  OB  =  OD,  the  square  of  BP  is  greater 
than  the  squares  of  OP  and  OB  by  the  rectangle  by  which 
the  squares  of  OP  and  OD  are  greater  than  the  square  of  DP 
{Euc.  IL  12,  13) ;  hence 

Bp2  ^  j)p.  ^  Qg.  ^  Qj)2  _^  gOP"  is  constant : 

similarly,  AP^  +  CP-^  =  AO'^  +  OC'  4-  20P'  is  constant, 
therefore  also    AP"'  +  BP'^  +  CP'  +  DP"'  is  constant. 

3.  (a).  Let  B  (fig.  57)  be  any  point  in  the  circumference  of  a 
circle  whose  centre  is  A ;  in  AB  take  two  points  C  and  D,  such 
that  ACAD  =  AB';  bisect  DC  in  E,  and  draw  EF  at  right 
angles  to  AE ;  in  EF  take  any  point  G,  then  will  the  tangent 
drawn  from  G  to  the  circle  be  equal  to  GO. 

Draw  GF  the  tangent  to  the  circle,  join  CG  ;  then 
AG^  =  GF^  +  AF'  =  GF'  +  AB', 
also   AG'  =  GE^  +  AE'  =  GE'  +  CE'  +  ACAD  {Euc.  ii.  6), 
=  CG^  +  AB^  by  construction ; 
...  GF'  +  AB^  =  CG'  +  AB', 
and  GF  =  CG. 

(/3).  Describe  a  circle  which  shall  pass  through  a  given  point, 
touch  a  given  straight  line,  and  cut  orthogonally  a  given  circle. 

Let  D  be  the  given  point,  BFL  the  given  circle,  KH  the 
given  line  intersecting  AD  in  H. 

In  AD  take  AC  :  AB  ::  AB  :  AD,  and  HK  a  mean  pro- 
portional to  HD  and  HC ;  the  circle  through  CD  and  K  shall 
be  the  required  ciicle. 


1851.]  EUCLID.  73 

For  the  centre  of  this  circle  will  be  at  some  point  G  of  EG, 
and  since  GF  =  GC,  will  pass  through  F,  cutting  the  circle  BFL 
orthogonally  at  that  point. 

Also  since  HK  is  a  mean  proportional  to  HD,  HC,  or 
HK"^  =  HD.HC,  HK  will  touch  the  circle  through  the  points 
C,  D,  K ;  hence  that  circle  fulfils  all  the  required  conditions  and 
is  the  circle  sought. 


(     74     ) 


ALGKBRA. 

1848. 

1.  A  walks  to  Trumpington  and  back  by  Granchester  In 
1^  hour,  starting  between  2  o'clock  and  2f ;  B  walks  the  same 
distance  in  the  same  direction  in  labour,  starting  between  2 
and  2^ :  find  the  chance  that  A  overtakes  B  before  he  gets 
home. 

Unless  A  starts  after  B  he  cannot  overtake  him. 

Now  if  he  starts  between  2  and  2^,  it  is  an  even  chance 
whether  he  starts  first  or  not ;  otherwise  he  cannot.  Hence  ^'s 
chance  of  starting  before  B  =  ^.|  =  ^. 

Again,  A  gets  home  between  3^  and  4,  B  between  3^  and  3f ; 
therefore  A  has  an  even  chance  of  getting  home  first ;  therefore 
also  his  chance  of  getting  home  last  =  ^,  and  chance  of  his  not 
overtaking  B  =  chance  of  his  starting  first  +  chance  of  his 
getting  home  last 

=  ^  +  ^  =  1; 

therefore  chance  of  A  overtaking  -C  =  1  —  f , 

the  required  chance. 

2.  A  paralleloplped  is  cut  by  three  systems  of  parallel  planes 
given  in  number,  parallel  to  the  three  pairs  of  opposite  faces 
respectively :  find  the  total  number  of  parallelopipeds  formed  in 
every  way. 

Let  in,  w,  j9,  be  the  given  number  of  intersecting  planes 
parallel  to  the  three  sides  respectively;  we  thus  have  m  +  2, 
«  -f  2,  ^9  +  2  parallel  planes  in  -three  several  directions. 

Now,  out  of  the  first  set  of  parallel  planes  we  may  make 

^ '— — — -  sets  of  two  each.     Similarly,  out  of  the  other  two 

sets  we  may  make  — ^-^ — —  ,   — ~- sets  respectively. 


1848.]  ALGEBRA.  75 

Now  each  paralleloplped  is  formed  by  taking  one  out  of  each 
of  the  above  sets  of  two  parallel  planes,  therefore  the  total 
number  of  parallelepipeds  will  be 

{m  +  2){m+l)     {n+2){n  +  l)    (p  +  2)(p+l) 
2  '  2  •  2  ' 

_  {m  +  1) {n  +  l){p  +  1) [m  +  2) (n  +  2)  (^  +  2) 
8  ' 

the  required  number. 

3.  If         {^-a){i/-ma)  =  {n^-mo(.){x-a) (1), 

and    O'  -  a')  (y  -  m/3')  =  {m^'  -  na)  (cc  -  /3') (2) , 

shew  that    [—^-^-\x  = -, ^-^r-  ■ 

\aa       pp  J  aa  pp 

Taking  (1)  {/3'  —  a)  —  (2)  (/3  —  a),  so  as  to  eliminate  y,  we  get 

m{^-oi){/3'-a'){^'-a)={{n/3-ma){l3'-0L)-{ml3'-na')[^-a)]x 

-  (n^  -  ma)  (^'  -  a')  a  +  (?n/3'  -  na]  [^  -  a)  ^8', 

or  m  {(/3 -  a]{l3"'  -  a  13')  +  [^'  -  a:)[oi' -  ayS)}  =  [n- m)(/3/3'  -  aa> 

-■n{(;S'-a>^+(;S-a)a'/3')l+m{(^'-a>'^+(/3-a)/3"0], 

.-.    {n-m){  (/3'  -  a')  a/3  +  (/3  -  a)  a^']  =  (h  -  w)  (/3/3'  -  aa')  a-, 

and  dividing  by  aa'/S/3', 

/J l_\  ^  g  +  g'  _  /3-F  /3' 

"^Vgg'      m'J        da  /3/3'    ' 

4.  (g).  Shew  that  the  integral  parts  of  (3*  +  1 )'""''  and 
(3*  +  1)'^'"  +  1  are  respectively  divisible  by  2'"'^^  where  m  is 
any  integer  whatever. 

The  integral  part  of  (S^+ir^^  is  (3*  + If'""^  -  (3*- 1)"""S 
since  it  is  a  whole  umnber,  and  (3*  —  1)^'"+^  is  less  than  1. 
Now  generally 

^.«.,  _  y-.^.  =  [x-y)  [[x^'+fl  +  try  (a;-*- ^4/"'-0 

+  a^/(a.— *  +  yn  +  ...  +  x'f (A), 

and  if  £c  =  3*  +  1,     y  =  3*  -  1, 
x-y  =  2,     xy  =  2, 
.-.    (3*  +  1)"'""''  -  (3*  -  1)'^'"*'  =  2  [{(34  +  l)'-""  +  (3*  -  1)'""} 
+  2  {(3*  +  l)"^"-^  +  (3i  -  I)"-""-'''}  +  . . .  +  2"']. 


76  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

This  part  of  the  question,  then,  reduces  itself  to  shewing  that 
(3i+  1)'-""  +  (3*-  l)'-""  is  divisible  by  2"'. 

Again,  the  integral  part  of  (34  +  l)'''"'  +  l  is  {Si+1)''"'+{S^-1)'% 
since  this  is  a  whole  number  and  (3*—  1)^'"  is  less  than  1 ;  hence 
the  second  part  of  the  question  reduces  itself  to  shewing  that 
(34+1)'"'  +  (3i-  1)'^"*  is  divisible  by  2"'^\  and  therefore  mcludes 
the  first  part. 

Now  (3*  +  1)*"  +  (3*  -  1)""  =  (4  +  2.34)'^"  +  (4  -  2.34f ", 

=  2''"'{(2  +  34)'-'"+  (2-34n, 
which  is  evidently  divisible  by  2^"^\     Also  generally 

_4«+2      I  4'i+a  /     2     1         2\   f  /     4»     ,         4»\  2     2/     4«-4     .         4n-4\ 

+  a.yK- +/"-), 

-...+  (-)VY"}; 
.-.   (34  +  1)*"+'''  +  (34  -  1)*'^'^  =  {(34  +  l)'-'  +  (34  -  1)'} 
[{(34  +  1)*"  +  (34-  1)*"}  -  2'-^{(34  +  l)*«-^+  (34-1)^"-^}  +...+  {-Y2'"l 

which,  since  (34  +  1)'"  +  (34  -  1)*"  is  divisible  by  2'"^',  is  divisible 
by  2'"+",  and  therefore  a  fortiori  hy  2'"'"+'  or  2^''''*'^*\. 
Hence,  whether  ?n  be  of  the  fonn  2n  or  2n  +  1, 

(34+1^"+  (34 -If" 

is  divisible  by  2'"^*,.  and  both  parts  of  the  proposition  are  true. 

(/5).    Prove  that  for  a  given  integral  value  of  a,  there  are 

(1),  a  integral  values  of  b  which  will  make  the  integral  part 
of  (a  +  b'-Y'"^'  divisible  by  2'"+^ ; 

(2),  a  integral  values  of  b  which  will  make  the  integral  part. 

of  {a  +  b'~Y"'^'  +  1  divisible  by  2'"^^ ; 

(3),  2a  integral  values  of  b  which  will  make  the  integral  part 
of  {a  +  UY"  +  1  divisible  by  2'"+\ 

(1).  The  integral  part  of  (^4+  af'*'  =  (&4+  af"^'  -  [U-af'^-'j 
provided  b^  —  a  <  I  and  >  0,  i.e.  if  b  lie  between  a^  and  [a  +  1)"^, 
which  gives  only  2«  values  of  b. 


1849.]  ALGEBRA.  77 

Now  by  equation  (A), 

+  [h  -  «^)  [{b'-  +  ay-'-'  +  {hi  -  aY"'-'}  +...+  (/._  aTl 

Now  since  b  is  only  to  have  (a)  values,  we  may  make  b  —  (i^ 
even,  and  then  the  problem  is  reduced  to  shewing  that 

{bi  +  ay"+{bi-af" 

is  divisible  by  2",  which  will  h  fortiori  be  tme  if 

{b'-  +  ar  +  {bi-aY% 

{the  integral  part  of   {b^'  +  af'  +  1}    be   divisible   by  2"^',   and 
therefore  (1)  reduces  itself  to  (3). 

(2).  The  integral  part  of  {a+b^-f"'^'+l={a+b^-Y"'^'+{a-biy"'^\ 
provided  a  —  b^  <  1  and  >  0 ;  therefore  b  must  lie  between  n^ 
and  [a—  1)'\  and  it  will  appear  by  a  precisely  similar  process  to 
the  above  that  (2)  reduces  itself  to  (3). 

(3).  The  integral  part  of  (a  +  b^Y'"  +  1  =  {a+b'-f"'  +  {a  -  7>i)*", 
provided  a  —  b^<  1  and  >  —  1,  i.e.  if  b  lie  between  [a—  1)'^  and 
{a+  ly,  giving  only  (4«)  admissible  values  of  b. 

And  if  we  flirther  make  b  —  d^  or  d^  —  h  even,  the  number 
of  values  of  b  will  be  reduced  to  (2a) . 

Then,  these  conditions  being  satisfied,  (3)  can  be  proved  as  in 
(a),  only  writing  5*  for  3*  and  a  for  1. 

Hence  the  propositions  enimciated  are  tnie. 

1849. 

1.  A  quantity  of  com  is  to  be  divided  amongst  n  persons, 
and  is  calculated  to  last  a  certain  time  if  each  of  them  receive 
a  peck  every  week ;  during  the  distribution,  it  is  found  that  one 
person  dies  every  week  and  then  the  coni  lasts  twice  as  long  as 
was  expected :  find  the  quantity  of  com  and  the  time  that  it  lasts. 

Let  X  =  the  number  of  pecks  of  corn, 

y  =  weeks  it  is  expected  to  last, 

then    -  =  the  whole  number  of  persons  =  n ; 

y 

or   X  =  ny (!)• 


78  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1H49. 

Also  n  =  number  of  pecks  distributed  in  tlie  1st  week, 
V  —  1  =  2nd 


n  -  r  +  I  =  J-"" 


and  it  lasts  2i/  weeks ;  therefore 

n  +  {n  —  I)  +  {n  —  2)  +  ...  +  {n  —  2?/  +  1)  =  whole  quantity  of  corn, 

or    {{2{n-i/)  +  l}?/  =  x (2); 

.-.  by  (1)    2{n-y)  +  I  =  n, 

n  +  I 

_  7i[n  +  \) 
^  -         2        ' 

and   2y  =  n  ■\-  1, 

which  determines  the  quantity  of  corn,  and  the  time  it  lasts. 

2.  If  ,^,  C,.  be  the  number  of  combinations  of  m  things  taken  ?• 
together,  and  jy  be  less  than  m  and  ??,  shew  that 

We  have    (1  +  xY{\  +  .-r)"  =  (1  +  a-)'"^". 

Now  the  coefficient  of  £c"  in  (1  +  .t)"  is  vOfx'.  also  the  coef- 
ficient of  x^  in  (1  +  a;)"'"^"  must  be  the  sum  of  the  jiroducts  of  the 
coefficients  of  a;'  in  (1  +  .x)'"  multiplied  into  the  coefficient  of 
af""  in  (1  +  x)"  taken  for  all  values  of  r  from  0  to  j>.     Hence 

3.  If  ^Gr  be  the  number  of  combinations  of  n  things  taken 
r  together,  prove  that  if  a  be  an  integer  greater  than  1,  then 
will   „,(7,„  be  greater  than  („C,.)". 


1849.]  ALOEBILV.  79 

The  total  number  of  combinations  which  can  be  made  out 
of  ?w  things  is  „„C,.„.  But  if  we  divide  the  na  things  into  a  sets 
of  n  each,  and  restrict  ourselves  to  those  combinations  containing 
ra  things  which  can  be  made  by  takuig  r  out  of  each  set,  we 
shall  get,  since  ,,  (7,  combinations  of  r  things  may  be  made  out  of 
each  set  of  w  things,  [„C,.)"  combinations. 

Hence,  since  (7.„  includes  every  possible  mode  of  fonnationa, 
and  („6'^)"  only  one  particular  one,  it  is  clear  that 

4.    If  ^  be  greater  than  unity, 

1  1  1.2 1.2.3 


and  if  it  be  less, 
1  1  1.2  L^__ « 


1-j,      1  +i?      (1  +p)  (1  +  2p)^  ^  (1  +p)  (1  +  22j)  (1  +  32)) ■ 

We  have  in  general,  when^  is  greater  than  unity, 

1 \__  _  M  +  1        1 

p  —  \      ^;  +  w      2^  -\r  n' 2^  —  X"* 

1  1  ??  +  1        1 


J)  —    ^         J9  +  7?         ^?  +  7i  '/>  —   1  ' 

therefore,  putting  successively  n  —  1,  2,  3 , 


112  1 

+ 


jj  —  1     2.)  ■\-  \     ^  +  1'^>  —  l' 


1  +^f^+ 


^j  +  1      ^^  +  1   V7?  +  2       y>  +  2    ^  -  1 
1  1.2  1.2.3 

+  T— ^T7 n^N   + 


■^  +  1  "^  (^+l)(^,  +  2)  ^  (^>+l)(7.-+2)  •;>-!' 

1  1.2  1.2.3 

+  7 — r-^, — r^x  +  7 — ,  ,x/  ,  ■  nN/..  ■  ox  +•• 


2>^\       (i?+l)(i^  +  2)  "  (^j+  l)(iJ  +  2)(^^  +  3) 


80  SOLUTIONS   OF  SENATE-HOUSE   PKOBLKMS.  [IHot*. 

Kj)  be  less  than  unity, 

_^ I  _  ^  {n+_l)p    _i_  . 

1  —J)       1  +  np        1  +  lip   '  1  — /' ' 
therefore,  putting  ?«  =  1,  2,  3,...  successively, 
I      _      1  2p  1 

1  -i>  ~  1+p      l+p  '  \-p  ' 

-      ^  2p     /      1  3/>  1    \ 

~  1  +/?''"  1  +p  li  +  2^ "''  r+ 2p  •  1  -pj ' 

~  l+p'^  [l+p)[l^2p)^'  "^  (l+^)(l  +  22^)(l  +  3/^)^' 

-f  

5.  A  bag  contains  three  bank-notes,  and  it  Is  known  that 
each  of  them  is  either  a  £5,  £10,  or  £20  note ;  at  three  suc- 
cessive dips  into  the  bag  (replacing  the  note  after  each  dip)  a 
£5  note  was  drawn :  what  is  the  probable  value  of  the  contents 
of  the  bag  ? 

There  are  six  possible  states  of  the  bag,  viz. 

(1).    3  £5  notes  in  which  case  the  value  would  be  £15. 

(2).    2  £5  and  1  £10 £20. 

(3).    2  £5  and  1  £20 £30. 

(4).    1  £5  and  2  £10 £25. 

(5).    1  £5  and  2  £20. £45. 

(6).    1  £5,  1  £10,  and  1  £20 £35, 

and  these  are  all  a  priori  equally  possible. 

Now  the  chance  of  the  observed  event  in  case  (1)  is  1, 

(2)  or  (3)  (ror|, 

(4)  (5)  or  (6)   (i^or^, 

therefore  the  probable  value  of  the  contents  is 

1  X  £15  +  A (£20  +  £30)  +  sV(£25  +  £15  +  £35) 
1  +  2  X  I,  +  3  X  i 

—    *         ~        4fi  1 

27 

_  4?nio 

—  ^mi 

=  £19  15.?.  V^^d. 


ALGEBRA.  81 

1850. 

1.  Prove  that  the  sum  of  the  fractions  which  are  intermediate 
in  magnitude  to  any  two  nmnbers  vi  and  n,  and  liave  3  for  a 
denominator,  is  n'  —  ni^. 

The  fractions,  together  with  the  intermediate  whole  numbers, 


will  be 


3w  +  1       ^m  +  2  3n  -  1 

3^"'  3       '  3 


.      /3m  +  1       3«  -  1\    (3^2-1)  -  3»« 
whose  sum  is    I — 1 — — 1 , 

_  [m-\-n)  [^[n-m)  -  1} 
_  _  ^ 

and  the  sum  of  the  intermediate  whole  numbers  is 

(m+1) +  (?»  + 2)  +...+  («- 1), 
»  —  1  —  m 


=  [[m  +  l)  +  [n-l)^ , 

(■w  +  r?)  {ii  —  m—  1) 
^  2  ' 

therefore  the  sum  of  the  fractions  is 

(w  +  n)  (3  [n  —  m)  —  1}       {m  -f  n)  [n  —  m  —  1) 
2  2  ' 

=  n'  -  m\ 

2.    There  are  a  number  of  comiters  in  a  bag,  of  which  one  is 

marked  1,  two  marked  2, up  to  r  marked  r ;  a  person  draws 

a  coimter  at  random,  for  which  he  is  to  receive  as  many  shillings 
as  the  nmuber  marked  on  it :  find  the  value  of  his  expectation. 

Since  the  person  is  as  likely  to  draw  one  counter  as  another, 
the  value  of  his  expectation 

total  value  of  contents  of  bag 

number  of  counters  in  bag    ' 
1'^  + 2^ +  ...+  ,.« 
=    l  +  2+...+  >-    '^'^^'''^'' 


82  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 


Now  I'  +  2'  +  ...  +  r'  = „ , 


o'{r  +l)(2r+l) 
6 


r(r  +  l) 
1  +2  +...  +  r=     \^     '  \ 

r(r  +  l)(2r  +  l)       r(r  +  l) 
hence  his  expectation  =  -=^ '-^ 1 r — - 

=  — - —  shillings. 


3.   If  a,  J,  c  be  in  harmonic  progression,  shew  that 

111  1         ^ 

-  +  -  + 7  + 7  =  0. 

a      c      a  —  b      c  —  o 

T    *  7>         1  1  1 


>S'  /3  +  a'  /3-a' 

then    -  4-  -  =  2/3, 
a      c 

and   a  -  b  =  -p: -^ , 

_  a 

-"/3(/3  +  a)' 

7.  1  1 


_        a 

"/3(/3-a)' 
^._  _J_  _^  _J_  ^  /3(^-a)  _  /3(/3  +  a) 
'  '  a  —  b      c  —  b  a  a        ' 

=  -  2/3, 
1111.. 

.-.     -  +  -+   7   +   7=0. 

a      c       a  —  b      c  —  b 

4.  If  there  be  z  counters  of  which  z  are  marked  m\  z  .n,... 
with  or  without  other  marks;  2!„,^,  w,  w,  with  or  without  other 
marks;  2;^,,„  ,,,y...  marked  w,  w,  ^j,  ^...;  the  number  mimarked 
is  s  —  22 „+  22;,„,„—  22;„,,„,p+...,  2  involvmg  all  combinations. 


1850.]  ALGEBRA.  83 

Let  Z>,,,  denote  the  operation  of  selecting  from  the  counters 
those  marked  with  ??«,  D^  those  marked  with  w,  &c.  Then  it  is 
manifestly  the  same  thing  whether  we  first  select  from  the  heap 
those  marked  ;/?,  and  then  from  these,  those  also  marked  n  ;  or 
whether  we  first  select  those  marked  w,  then  from  these,  those 
marked  m ;  or  at  once  select  those  marked  7n,  n.  Tliis  may  he 
symbolically  expressed  thus : 

D  D  =  D  D    =  D     ; 

similarly,  we  have  in  general 

D  D  D  ...  =  D  D  D  ...  =  ...  =  D       ...:  (1). 

Also  1  —  Z),,,  will  denote  selecting  those  unmarked  with  m^ 
1  —  D^^  those  immarked  with  ??,  &c.  Hence  the  whole  number 
unmarked         _  ( i  _  2)  j  (i  _  i)J . .  ,z^ 

=  (1-22)   +2i)      -2i>,,    „+...)z', 

since  the  spnbols  D^^^^  A.v  have  been  shewn  (equation  (1))  to 
be  commutative 

■^    m     '     ^    m,n  ^    m,n,p  ^•••1 

the  required  number, 

5.    Prove  that 

a;'  +  /  +  (a;  +  y)«  =  2  (.x"  +  xy^-  ?/)*  +  SxY [x  +  yf  (ar*  +  xy  +  f] ; 

and  if  x^  +  xy  -{•  y^  =  a,  xy  {x  +  \j)  =  i,  and   n  be  any  positive 
integer,  shew  that 

^^»+y^-+(^+^)^»^2a"  +  n(n-2)a-6-  +  "("-'^^;;;'^^"-'^a-^- 

/^(»-r-l)(7^-r-2)...(ri-3r  +  l)     ,_„.^,. 
3.4. ..2r 

(a).    Let  2  be  a  quantity,  such  that 
X  +  y  +  z  =  0', 

then   x""  +  f  +  [x^-yT  =  x'"  +  /"  +  z'\ 

g2 


84  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

and   x''  +  xy  +  y^  =  {x-\-  yj  —  xy^ 

=  -z{x  +  y)  -xy, 
=  -  {yz  +  zx  +  xy)j 
xy{x  +  7/)  =  -xyz- 
therefore,  taking  the  notation  of  the  latter  part  of  the  question, 
yz  +  zx  +  xy  =  —  a, 
xyz  =  —  by 
therefore  x,  ?/,  s,  are  the  roots  of  the  equation 

.•.    {^  —  x){^  —  y){^  —  z)  =  ^^  —  af  +  hj  identically j 

-i)(-D(-i)--M' 


l-|.(«-jy, 


.-.  log(l-|)+log(l-|)+log(l-|)=logjl-^,(a-| 


x  +  t/  +  z  ,   1  a;'  +  /  +  ^'  ,       ,    1    x"  +  /"  +  z'"  , 
••  f         +2  f  ''"••■"'"2m  f"  "'"• 

1  /       6\        1    /       i\^'  1    /       i\" 


=  rr-|;  +  2rr-|j  +-+s|-^«-|j+-; 

therefore  equating  coefficients  of  ^j , 

i_  (aj--"-  +  y2»  +  2'-*«)  =  1  a"  +  -^  {n-l){n-2)  ^„^,^, 
2n  n  n—1  1.2 

J_  (n-2)(n-3)(n-4)(7^-5)     , 
■^71-2  1.2.3.4  ""    ^  +••• 

1      [n-r){n-r-l)...{7i-?,r+\) 
»i  — r  1.2. ..2r 

n  2  2.3.4 


(9i-r-l)...(n-3r+l) 
^  2.3. ..2«  "*      ^    +• 


1850.]  ALGEBRA.  85 

.-.    x'"  +  /"  +  [x  +  T/y  =  2a"  +  n{n-  2)  a"-'b' 


{n-S){n-A){n-5)  ^,.^^,  _^^ 


n 
+  -  3.4 


n{n-r-i)...{7i-Sr+l)    ^ 
^  3.4.. .2r  ""      ^    ■^•••• 

Hence  putting  n  =  4,  we  get 

a;'  +  y  +  (a;  +  yY  =  2  [x"  +  xij  +  ff  +  S{x''  +  X7/  +  f)  xY  {x  +  yf. 

/aV'^'' 
6.    (a).  If  a  be  less  than  h,  prove  that  ij-j      is  increased  by 

adding  the  same  quantity  to  a  and  b. 

(/3).   And  if  w  be  greater  than  1,  shew  that 


by  means  of  this  fonnula  prove  that  ^ 

(a^  +  a,  +. . .+  a  J"  >  n\a^. .  .a^. 

(a).  Since  a  is  less  than  &,  we  may  put  a  =  h  —  c,  where  c  is 
a  positive  quantity  less  than  b ;  then  a  and  5  will  be  increased 
by  the  same  quantity  if  we  increase  Z>,  c  remaining  constant  j 

^y      "V~^j       =  a:  suppose; 


•.  loga;=  {2b-c)\og(^l  -^  , 


/         c\  /,       1  c       1  c' 


2      l\c      /2      l\c'  /2         1 


ti-i 


which  is  manifestly  negative,  since  the  coefficients  being  of  the 
form  ( 


-  j  are  positive ;  and  the  absolute  magnitude  of  the 
series  is  diminished  by  increasing  b  if  c  remain  unaltered ;  hence 

(      V  0+6 
r)     ,  is  increased  by  adding  the  game 

quantity  to  n  and  h. 


86  SOLUTIONS    OF    SENATE-HOUSE    PROBLEMS.  [1850. 

(/3).*    First,  let  J  be  greater  than  unity  =  1  +  -  suppose; 
we  have  then  to  shew  that 


nj         L         n—1  J  V         "  —  1 , 

these  quantities  when  expanded  by  the  Binomial  Theorem  become 

,  +  ^  +  __^'  + —i *+•■■' 


,                            n  —  \     .,       V         n—\j\         M—  1/      3  . 
and    1  +  a;  +  — 7-T x-  + -—- x  +... ; 

all  the  terms  of  both  series  are  positive,  and  each  term  of  the 
first  series  greater  than  the  corresponding  term  of  the  second,  or 

1  +  7J  "(^  +  ^1)  '^^^  U)  >toj  ' 

when  a  is  greater  than  h. 

Cb  .  X 

Next,  suppose  j  less  than  unity  =  1  —  -  suppose,  we  have 
to  shew  that 

l-i^^    >    J % or    1  =" 


w/         I         w  —  1  J  V         n  —  1 

(.  -ii-i 
1 j        >  (1  —  x)''": 

these  quantities  when  expanded  by  the  binomial  theorem  become 


71  —  \       „  V  71—  l)    \  71—  \, 

1  +  ^^+      1.2      ■'"  + 1:2:3 ^  +■■•' 

7^       .y  \  71      \  71/        ., 

and  1  +  X  +  ^^  a.^  + ^-^^^ ^^  + . . . ; 

*  'This  part  of  the  solution  is  given  by  Mr.  Thacker  in  a  recent  number  of 
the  Vamhridfie  and  Dublin  Mathematical  Journal,  No.  xxv.  p.  8L 


1850.]  ALGEBRA.  87 

all  the  terms  of  both  series  are  positive,  and  each  term  of  the 
first  series  greater  than  the  con'esponding  term  of  the  second, 
and  therefore  the  proposition  is  true  in  this  case  also. 

Now  let  the  n  quantities  «j,  a^v^n?  ^®  ^^  ascending  order  of 
magnitude ;  then 

V  n  /  '    V  na^  )  ' 

^"^    1'+  [n-l)a,  I     ' 

by  the  above, 

^rt.  -I- «,+...+ a 


n-l 

>  a 


similarly  (  -^ 5__^ « j      >  a_^  f  _! *  " 


w-l  '      ' 

n-i  f  ^     I    ^      ,  I    ^  \  n-2 

? 


> 

n-l    '       H 


,   >  a   ,a  : 


hence  by  multiplication 


or  [a^-\-a^^-..,+  aJ'  >  n{a^a^..,aj. 

7.   If  -  be  the  r'^  fraction  converging  to  — ,  and  n'  be  the  r*^ 

remainder  in  the  process  for  finding  the  successive  quotients, 
prove  that 

m     p  _  p 

n       q      pq ' 

Let  ^  ,  ^  be  the  {r  -  2)'^  and  (r  -  1)*  converging  frac- 

9'r-2       2'r-l 

tions  respectively;  m  the  r'^  quotient,  n'  the  (r— !)'•»  remainder; 

then  i^   =   »'^>r-,    +i^r-25 


88  SOLUTIUNS    OF   SENATE-HOUSE    PKUBLEMS.  [1850. 

And  the  fx"actioii  —  may  be  derived  from  -  by  wTitiug  rii-\-  —r, 
for  111  In  the  above  expressions  for  p  and  q ; 


in 


'«' +  :^' )  i^r-i  -^Vr^ 


Now  —  is  in  its  lowest  tenns,  and  n  is  prime  to  n", 

.-.    m  =  [m'n" ■^n)p^^^  +  n"p^_^^ 
n  —  [m'n"  +  n)  q^._^  +  n'q^_,^^ 


,    m      p      mq  ^  np 

anci        '^      -^ . 

n       q  nq 

Jig-  * 

8.  Find  the  probability  of  drawing  a  black  and  a  white  ball 
the  same  number  of  times  from  a  bag  which  contains  an  equal 
number  of  each ;  the  balls  being  dra^\ai  one  by  one  and  replaced 
after  each  draAving,  and  the  nmnber  of  drawings  being  the  same 
as  the  number  of  balls  in  the  bag,  but  this  number  is  unknown, 
any  number  from  2  to  2n  being  equally  probable. 

Suppose  there  are  2x  balls  in  the  bag;  the  number  of 
drawings  will  then  be  2a;,  and  the  number  of  possible  ways  in 
which  the  balls  may  come  out  =  2'''''". 

Of  these  the  number  of  favourable  cases  equals  the  number 
of  permutations  of  2x  things  taken  all  together,  of  which  x  are 
of  one  kind  and  x  of  another, 

_   1.2. ..2a; 

~(l.2...a;y^' 

_  (a; +1) (a; +  2)... 2a; 

~  1.2. ..a;  ' 

therefore  chance  of  proposed  event  on  this  supposition 

_  J_  (a- +1) (a; +  2)... 2a; 
~  ^r  1.2. ...r 


I851.J  ALGEBRA.  89 

Now  the  chance  of  there  being  2x  balls  in  the  bag  =  -   what- 

n 

ever  number  >  0,  >  «,  x  may  be.     Hence  the  chance  of  tlie 

proposed  event 

^    ^^   !^l!i   .        .1    in+l)in  +  2)...2n 


n   (2M       2M.2  2'"'  1.2 

1851. 

1.    If  y  ,  y7  be  fractions  in  their  least  terms,  the  denominators 

of  which  do  not  exceed  a  given  nmnber  ??,  the  fonner  fraction 
being  given,  and  the  latter  detennined  from  it  by  taking  for 
a  and  b'  the  greatest  values  of  x  and  y  (?/  not  greater  than  w) 
which  satisfy  the  equation  hx  —  ay  =  1,  then  of  all  the  fi'action 
in  their  least  tenns,  the  denominators  of  which  do  not  exceed  w, 

the  fraction  j-,  exceeds  y  by  the  smallest  quantity. 

,,,.    ,  a       a       ha  —  ah'        1 

Wehave  _  _  _  =  _^^  =  _  , 

since  a',  h'  are  values  of  x  and  y  in  the  equation  hx  —  ay  =  1. 

Let  ^  be  any  other   fraction  in  its  least  tenns  whose  de- 
nominator does  not  exceed  /i,  then 
a       a       ha  —  a/3 
^~h^  ~ir^       ~  h^ 

m  being  some  integer  greater  than  1 ;  then  a  and  ^  are  values 
of  X  and  y  in  the  equation 

hx  —  ay  =  m. 
Now  this  equation  is  satisfied  by 

X  =  ma  ±  qa^ 
y  =  mh'  ±  qh^ 
where  q  is  any  integer. 

Again,  the  successive  values  of  x  and  y  which  satisfy  the 
ccpiation  ;,,y  _  ^,y  ^  1^ 


=  Ta  say, 


90  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

differ  by  a  and  b  respectively ;  hence,  as  b'  is  the  greatest  value 
of  y,  less  than  n,  in  this  equation,  b'  +  b>  n,  therefore  h  fortiori 
mb'  +  b>  n.  Hence  /S  cannot  be  of  the  form  mb'  +  qb^  since 
it  is  less  than  n. 

Neither  can  it  be  of  the  fonn  mb' ;  for  then  a  would  =  ?««', 

and  p  would  not  be  in  its  least  tcnns. 

Hence  yS  must  be  of  the  form  mb'  —  qb : 

.•.  /36  =  [mb'  —  qb)b  <  mbb'j 


a 

a        m          1 

13 

b  ^  mhb'  ^  bb' ' 

a       a 

or  of  all  the  fractions  in  then*  least  terms,  whose  denominators 

t 

do  not  exceed  ??,  y?  exceeds  r  by  the  smallest  quantity. 

2.   The  sum  of  the  series  -^  +  -^^  +  -^^^  +  ...  to  infinity, 

where  S  is  positive,  is  greater  than  (2^  —  2)  "^  [^^  ~  ^)i  ^^^  1®^^ 
than  2* -=-(2^-  1). 

Let  8  be  the  sum  of  the  given  series. 
Then 

1         1  /_2_       ^       _8_ 
>  p+3  +  2  V2^*  "^  41+a  "*"  8^3  "*■ 

^  P  "^  2  V2^  ^  4^  "^  8"^  "^ 
>lH-i        ' 


2   2^-1 


2«-  1 


1851.]  ALGEBRA.  91 

Again, 

1         /J_       J_\        /J_        _1_       _1_       Jl_\ 

1         1         1 

^  jd  ^  2'»       4* 
1 

< 


2«-l 


2^  —  1  2^ 

Hence  S  lies  between  -i — ~-  and 


2*  -  1  2^  -  1  * 

3.    Solve  the  equation 

1  1  —  X  I  I  -  a      ^ 

X  + a 1 =  0 ; 

1  —  X  X  1  —  a  a 

and  thence  infer  the  resolution  of  the  first  side  of  the  equation 
into  factors. 

We  see  at  once  that  the  given  equation  is  satisfied  by  a;  =  a. 

Again,  for  a  write ; ,  then 

1      _  1  _  _  1  -a 

1  -  a  ~  1       ~      ~~ar  * 

I  —  a  1  —  a  , 

= • =  -  a. 

a  1 


1  -a 

Hence  the  given  equation  becomes 

1  1  -  X        ,  1  \  -  a       ^ 

X  + a  - 7  +  — —  =  0, 

I  —  X  X  1  —  a  a 

of  which  a  or    is  a  root. 


92  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Similarly,  ; —  is  a  root  of  the  given  equation. 

But  a_-l  a  1 


a  -  \  1  -  a  ' 


a 
therefore  the  roots  of  the  given  equation  are  a, , —  ; 

and  the  first  side  may  therefore  be  put  into  the  form 


x[l  —x) 

4.    From  the  equation 

ah[c  +  d-e  -/)  +  cd{e  +/-  a-h)  +  ef[a  +  h-c-d)  =  0, 

detennine,  in  terms  of  &,  c,  (7,  e,y,  the  ratios  a  —  c  :  a  —  d,  and 
a  —  e  :  a—f\  and  shew  that  the  relation  between  the  six  letters 
may  be  expressed  in  the  form  P  =  Q^  where  P  and  Q  are  each 
of  them  the  product  of  three  differences  of  pairs  of  letters,  or 
in  the  form  R=  8^  where  R  and  S  are  each  of  them  the  product 
of  four  differences  of  pairs  of  letters. 

(a)    From  the  given  equation  we  get 

_  cd[h -  e -/)  +  ef{c  +  d-b)  ^ 
^~    b{c  +  d-e-f)  -  cd  +  ef  ^ 

_  c[b-d)  (e+f)  -  he'  +  ef{d-b)  +  6'd 
•'•  ""      """  b[c+'d-e-f)  -cd  +  ef 

^       "^1  b{c  +  d-e-f)-cd^€f 

=.-(b-d] ('-')i'-f) (1) 

^  '  b{c  +  d-e-f)  -cd+ef  ^^' 

Snmlarly,    a-d=-{h-e)  ^^  f^^^^_^_^  J  ^^  ^y  5 


—  c  _b  —  d   c  —  e    c  —  f 

-  d      b  —  c'  d—  e'  d  —  f 


(2). 


1851.]  ALGEBRA.  53 

In  the  same  manner  it  may  be  shewn  that 

a  ~  e  _h  -  f    e  —  c    e  —  d  ,, 

^irj-h^e'T^c'J^d ^^^' 

(yS)    From  (1)  we  see,  by  interchangiug  c  with  e,  d  with  /, 
which  does  not  alter  the  original  equation,  that 

^      ^  '  l>[e-\-f-c-d)  -i-  ef-  cd^ 
a  —  c  _      h  —  d    {c  —  e){c  —f)  ^ 
'  '  a  —  e  ^  ~  f    (e  —  c)  (e  —  c?)  ' 

•••  («-«)  (*-/)  {e-d)  =  [h-d)  {c-f)  {a-e), 
which  expresses  the  relation  between  the  six  letters  in  the  fonn 
P=Q. 

(7)    Again,  by  (2)  we  get 

(«-c)  [b-c)  {d-e)  (d-f)  =  {a-d)  [h-d]  (c-e)  {c-f), 
which  is  in  the  form  B  =  S. 

5.    If  d^  + 1  be  exactly  divisible  by  j),  and  —  be  converted 

into  a  continued  fraction,  until  two  consecutive  reduced  fractions, 

— ,  — , ,  are  found,  such  that  p^  >  n  <  n.  then 
n  ^   n  '  ^  ' 

j^  =  [na  —  mjpf  +  w". 
It  is  manifest  that 

{na-mp)   =np  [-- -) 

.,  .,  (m       mV 

by  the  property  of  Continued  Fractions. 

-r,  m       m  1 

But  — =  +  — -, ; 

n        a  nn 

.'.  (na  —  mpY  <  ^, ; 

and  n''^  >  p ; 


94  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

therefore,  h  fortiori^  [na  —  mpf  <  p^ 

and  n^  <  p ; 

.*.   {na  —  inj})'^  +  n^  <  2_i). 

But     {na  —  mj)f  +  n^  =  n\d^  +  1)  —  27nna.p  +  m^p\ 

which  must  be  divisible  by  p,  since  a^  +1  is  so : 

and   {na  —  mpf  +  ri'  has  been  shewn  to  be  <  2^,  and  it  must 
be  positive ;  therefore  we  must  have 

{na  —  mpY  +  n^  —  p. 

6.    If  {a+yS+7+...}^  denote  the  expansion  of  (a+/3+7+...)", 
retaining  tho^e  terms  iVa"/3V^''-"  only  in  which 

&  +  c  +  ^+  ...  l!(>p—  1,    c  +  d+  ...  ■;^  p  —  2^  &c.  &c., 
then 

*This  theorem  may  be  put  into   a  rather  more  convenient 
fonn  by  wi'iting  x  —  a.  for  a- ;  we  have  then  to  shew  that 

{x  -  a.y  =  x''-n  {ay  {x  +  /S)"'^  +  ^ll^zil  («  +  ^Y  {x  +  ^  +  7)""'' 

or,  writing  a^  for  a,  a,^  for  /3,  &c., 

(a.  -  «,)•  =  a;"  -  « (a.)'  (*  +  «J-  +  '-^^^  (a,  +  «,f  (.r  +  a,  +  ocj""' 

-  "'""lilT"''  («.+«^+='.l'  (-+«.+«.+=•.)"-'  +  (1).    ' 

The  proof  of  this  depends  on  the  expansion  of  the  quantity 

{a, +«,+  ...  +  «,}". 

*  For  tlie  solution  of  this  problem  we  are  indebted  to  Mr.  Cayley. 


1851.]  ALGEBRA.  96 

Expanding  by  the  Binomial  Theorem, 

(a^  +  a,+  ...+a,)'' 

To  pass  to  {a,  +  a^  +  . . .  +  «/,}''.  The  sum  of  the  indices  of 
Og,  ttg,  ...  a^,,  are  not  to  exceed^  —  1  ...,  and  generally  the  sum 
of  the  indices  of  a,.,  a;.^^,  ...  a^,,  are  not  to  exceed  p  —  r  +  1. 
Hence  in  (a.^  +  ag  +  ...  +  a^)*",  the  required  conditions  will  be 
satisfied,  if  only  the  sums  of  the  indices  of  a^_,.+2,  ap_,.+3j  ■  •  •  «p  do 
not  exceed  *•  —  1,  and  the  sum  of  the  indices  of  a^.^^,,  a^,^^^, ...  a^ 
do  not  exceed  r  —  2,  and  so  on.  And  this  will  be  the  case 
if,  considering  a^  +  ag  +  . . .  +  a^_^+j  «^  one  quantity^  we  replace 
(a^  +  a3+  ...+a^)'by 

K^a  +  «3  +  •  •  •  +  Vm)  +  V-+2  +  •  •  •  +  a;,}^ 
Hence     {a,  +  a,+  ...  +a,|^  =  a,^  +  |a/"M(«2  + "3+  •••  V^+i^f)!' 
+  f  ^^  «/"'  {(a.  +  "3  +  •  •  •  +  a,_,.)  +  a^_^Y  +  ... 

+  pa^{a,  +  a^  +  ...  4  a,}""'  (2), 

the  last  term  (a.,  +  a^+  ...+  OLp)"  being  of  course  rejected  alto- 
gether, since  the  sum  of  the  indices  exceeds^  —  1. 

Now  the  coefficient  of  -   — - —  ... (-  a,)*",  on  the 

12  r 

left-hand  side  of  equation  (1),  is  cc"~'. 

On  the  right-hand  side,  it  is,  expanding  each  of  the  quan- 
tities in  the  brackets  ( }  by  (2), 

(aj  +  a,  +  ...  +  a,J"-'--^((a,+  ...  +  a,J]'(a^  +  «,  +  ...  +  a,,,)"-'-^ 


n-r-1 

r+2/ 


96  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [I8iil. 

or,  if  wc  write  y  fur  x  +  a.^+  ...  +  a,^,,  to  for  a.^  +  ...  +  a,,^,,  the 
coefficient  on  one  side  is  (^  —  w)""'',  on  the  other 

«-r  "  '      f       11/         ,  \M-'-l 

»  —  r  »  —  y  —  1  ,  ,,,  ,  Ml-,-!! 

+  —J 2 ^"^  "^  "''■^•'^    ^^  "^  ^'^'  ^  °''*='^        ' 

+  

Therefore  we  have  to  shew  that 

n  —  r  n  —  7'  —  1  ,  lo  /     ,  ,         \»i-)--> 

+  _ _ {a,  +  a  ,,r  {y  +  a .,,  +  a  ,3)"  "-... 

the  theorem  itself,  wiiting  n  —  r  for  n. 

The  coefficients  of  a^**,  on  each  side  of  the  equation  (1),  are 
obviously  equal.  If  then  the  theorem  hold  for  the  indices 
1,  2...(m  — 1),  it  is  proved  to  hold  for  the  index  n.  But  it 
obviously  holds  for  the  index  1 ;  therefore  it  holds  miiversally. 

7.   If =  a  +  — ^-— ,  and  — —, —  =  a  +    ,        , : 

mp  +  X  ]}  +  It  mj)  +  X  p  +  7^  ' 

then  supposing  ic  and  u  to  vanish  when  x  vanishes, 

1  1  .  ,s 

-7  —  =  m  (a.  ~  a.  . 
u       ti 

Since  u  and  u'  vanish  when  x  vanishes,  we  have 

1  B  1  .      )8 

—  =  a+— ,        — ,  =  a  +  —  , 
mp  J)  tnp  p 

Hence,  by  the  first  of  the  given  equations, 

p  +  u  1 

Tnp  +  X  m 

X 

.'.  p  +  u  =  au  {mp  +  a?)  +  ^  H 

oc 

=  w  ( 1  -  mB  +  oLx)  +  »  —  ; 


1851.]  ALGEBRA.  97 

.'.  u  (m8  —  ax)  =  —  , 
m 

m'/3  1 

.'. ma  =  -  . 

X  u 

Similarly,  by  the  second  of  the  given  equations, 

m^^  ,       1 

ma  =  —: 

X  u 


the  required  result. 


11, 

-, =  m  a  —  a 

u       u 


H 


(  ^'H  ) 


PLANE  TRIGONOMETRY. 

1848. 

A  cannon-ball  is  moving  in  a  direction  making  an  acute 
angle  6  with  a  line  drawn  from  the  ball  to  an  observer;  if 
V  be  the  velocity  of  somid,  and  nV  that  of  the  ball,  prove 
that  the  whizzing  of  the  ball  at  the  different  points  of  its 
com'se  will  be  heard  in  the  order  in  which  it  is  produced,  or 
in  the  reverse  order,  according  as  /?  <  >  sec  6. 

The  whizzing  will  be  heard  in  the  ordei'  in  which  it  is  pro- 
duced, or  in  the  reverse  order,  according  as  the  sound  or  the 
ball  moves  more  quickly  towards  the  observer.  Now  the  velo- 
city with  which  the  ball  moves  towards  the  observer  =  wFcos  d ; 
hence  the  whizzing  will  be  heard  in  the  natural  or  reversed 
order,  according  as 

V  ><  n  V  cos  0, 

or  no  sec  0, 

the  required  condition. 

„     j^  sin  (a  —  jS)  _  sin  (a  +  ^) 

sin  yS  sin  ^      ' 

shew  that     cot  /3  —  cot  0  =  cot  {ol  +  0)  +  cot  (a  —  /3). 


Since 


sin  (a  —  /3)       sin  (a  +  0) 
sin  /3  sin  0 

1  1 


sin  /3  sin  (a  +  0)       sin  0  sin  (a  —  yS)  ' 

sin  {a  +  0-^)    _   sin  (g  -  /3  +  6')  ^ 
sin  yS  sin  (a  +  0)      sin  0  sin  (a  —  /3)  ' 

.'.  cot  yS  -  cot  {a  +  0)  =  cot  0  +  cot  (a  -  /3), 
or  cot  13  —  cot  0  =  cot  {a  +  0)  +  cot  (a  —  /S), 

the  required  result. 


PLANE   TRIGONOMETRY.  99 

1849. 

1.    If  COS  a  =  cos  ^  cos  <^  =  cos  yS'  cos  0', 

and    sin  a  =  2  sin  ^(/> .  sin  ^0', 
shew  that   tan  \a  =  tan  \j3  tan  ^^8'. 

Since  sin  a  =  2  sin  \^  sin  |^<^', 

.-.    sin'  a  =  (2  sin'  :|<^)  (2  sin'  ^  <^') 

=  (1  —  cos  0)  (1  —  cos  0') ; 

,         /        cosa\   /         cos  a 
,1  —  cos  a  =    1 :x       1  - 


cos/Q/  V        cos/37  ' 
.'.  cos  a  =  sec  /3  +  sec  /S'  —  cos  a  sec  /8  sec  /8', 

,  sec  B  +  sec  yS'  cos  B  +  cos  /3' 

and    cos  a  =  :, ^^ 7^  = 7^ p=;  ; 

1  +  sec  p  sec  p       I  +  cos  p  cos  ya  ' 

2  <*  _  1  ~  cos  a  _  1  —  cos  y8  —  cos  /8'  +  cos  /3  cos  y8' 
2      1  +  cos  a      1  +  cos  /8  +  cos  /8'  +  cos  /8  cos  yS' 

^(l-cos^)(l-cos^;)  g         ff 

(l  +  cosy8)  (l+cosyS)  2  2  ' 

a  /8        yS' 

and  tan  -  =  tan  —  tan  —  . 
2  2  2 


2.    Find  •=•  from  the  equations 

[a  +  h)  sin  d  +  [a  -  h)  cos  6  =  {ci'  +  Jr)\ 

a  sin'  e  +  b  cos'  0  =  {Zah)K 

Squaring  the  first  of  the  given  equations,  we  get 

{(i'+V')  [mi'd^-  cos'^)  -  2ah  [cos'S-sm'S)  +  2(a'-  J')  sin^cos^=a'+  b'; 

.-.    (a'  -  &')  sin  2^  -  2rtZ>  cos  2$  =  0. 

Let  -  =  tan  </>,  then  this  equation  gives 

sin  2  (^  -  4>)  =  0, 

whence    2  (^  —  ^)  =  0  or  tt  ; 

.".    6  =  (f)  or  (f)  +  ^TT. 

11  -J 


100  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS,  [1849. 

Taking  ^  =  </>,  the  second  of  the  given  equations  gives 
a  siii^  <^  +  i  cos'  0  =  (3aJ)*; 


'fTI  +  ^7-^  =  ('«^^^-' 


a  J  \        a 

•  •  a  =  (3«&)*, 

and  72-3y+l=0; 
o  o 

«      3       1  ,       , , 
••■5  =  2±2(-')'  (')• 

Again,  taking  ^  =  ^+^7r,  the  second  of  the  given  equations  gives 

a  cos^  (f>  —  b  sin'*  <f)  =  (3aZ>)*, 

a  —  b  ,„   ,, , 

1  =  {^^^)H 

{a'-^bj 
.-.  d'-  W=  {a'  +  bf  {3abf, 
a*  -  Id'V  +  b^  =  Sab{d'  +  V'), 
(a^  +  yy  _  ^.d'b''  -  dab  (a'  +  J'^)  =  0 ; 

...     (a^  +  J2  _  4^J)    ^^2  _^  J2  ^  ^j^  ^  0 . 

therefore,  fii-st,  a''*  +  J^  —  4a5  =  0, 


^'Z           Kb. 
and  ^  =  2±(3)i (2): 


1850.] 

PLANE   TRIGONOMETRY. 

or,  secondly, 

d'  +  h'  -V  ah  =  0, 

and 

•••  ?=-i±i(-3)* 

101 


(3). 
The  six  values  given  by  (1),  (2),  and  (3),  are  all  the  values 

which  Y  admits  of. 
o 

1850. 

1.  If  through  the  angles  of  a  square  four  straight  lines  be 
drawn  externally,  making  the  same  angle  a  with  the  successive 
sides,  so  as  to  form  another  square,  find  its  area. 

Let  a  be  the  side  of  the  interior  square :  then  the  area  of  the 
exterior  square  =  interior  square  +  four  triangles  each  equal  to 
^d^  sin  a  cos  a,  or  =  d^  +  4  x  ^a^  sin  2a  =  d\l  +  sin  2a). 

2.  Shew  that 

2tan-Mtann45°-a)  tan  1/3}  =  cos"'  /J^^1^L±£^^ 
^^       ^  /        2A-J  Vl  +  tanacosyS;  ' 

if  a  be  <  45°. 

In  general    cos  (2  tan"' j:;)  =  2  cos^(tan~'cc)  —  1 

2 

1 


Let  X  =  tan*  (45°  — a)  tan^/3,  then 

^  ^       1  +  tan(45  -a)tan'^/3 

1  —  tana  1  —  cosyS 

_  1  +  tang  1  +  cos^ 

1  —  tana  1  -  cos/3 

1  +  tana  I  +  cos/3 

_    tana  +  cos/3 
~  1  +  tana  c<>s/3  ' 


102  SOLl'TK.lNS    OF    SHNATE-HuUlSE    rKOliLK.MS.  [I«o0. 

.-.  2  tan-'  {tan*(45"  -  a)  tan^/3}  =  cos-  (  tan«  +  cos^\ 
If  a  were  >  45°,  the  given  expressions  would  become  imaginary. 

3.  Draw  AB  and  AC  (fig.  58)  at  right  angles  to  one 
another,  and  make  AB  equal  to  twice  AC]  produce  CA  to  D 
until  CD  is  equal  to  CB:  prove  that  BB  will  be  the  side  of 
a  regular  pentagon  inscribed  in  a  cii'cle,  of  which  AB  is  the 
radius. 

Also,  if  with  centre  B  and  radius  BA  we  describe  a  circle 
AEF^  of  which  ABF  is  a  diameter,  and  make  AE  equal  to  AB^ 
then  FE  will  be  the  side  of  a  regular  pentagon  circumscribing 
a  circle,  of  which  A  C  is  the  radius. 

(a)    Let  AC  =  a,  then  AB  =  2a ; 

.-.   CB=CB=  [AB'  +  A  Cy  =  bhi ; 

.-.  AD=  (54-1)  a, 

BD  =  {AD^  +  ABy 

=  (6  -  2.5*  +  4)*a 

=  (10 -2.5*)*  a 

,     (10-2.5*)* 

=  4a  ^^ — 

4 

=  2  sini7r.2a 

=  2  sin  lir.AD ; 

therefore  i?i)  is  the  side  of  a  regular  pentagon  inscribed  in 
a  circle,  of  which  AD  is  the  radius. 

{^)    Again,     FE''  =  AF'-AE' 

=  {2.ADY  -  AB' 
=  4(6-2.54)rt"''  -  4a' 
=  (20  -  8.5*)  a' 
=  2^(5-2.5*)  [aY 
=  {2tsinl7ryAC'; 

therefore  FE  is  the  side  of  a  regular  pentagon  circumscribed 
about  a  circle,  of  which  AC  is  the  radius. 


1850,]  PLANE    TRIGONOMETRY.  103 

4.  If  (?,  hj  c,  be  the  sides  of  the  triangle  ABC,  2^1  ?>  ^  lines 
bisecting  the  angles  drawn  to  the  opposite  sides,  and  p\  g-',  r 
these  lines  produced  to  meet  the  circle  which  circumscribes 
the  triangle;  shew  that 

COsi^  COsii?  COsiC  111 

—    -I-    = —   -k^ = —  =  — I 1 — 

J)  ^"•""^  r  a       0       c 

p  cos^-4  4-  q  cob\B  +  r  cos^O  =  a  •\-  h  +  c. 

(a)  Let  AD  (fig.  59)  be  the  line  bisecting  the  angle  BAG, 
i>,  D'  the  points  in  which  it  meets  BG  and  the  circumscribing 
circle  respectively. 

Draw  DG,  DH,  perpendicular  respectively  to  AB,  AC] 
then  DG  =  DH  =  p  m\^A, 

and  D  G.AB  +  DKA  G=2  (area  of  triangle) 

=  AB.AC  sin  A', 

.'.  p  sin^-4  {h  +  c)  =  be  sin  J. 

=  be  2  sin|^^  cos^^  ; 

2  cos4^      1       1 

p  be 

.,.    .,    ,  2cosi^      1       1 

Similarly,  ^—  =  -  +  -  , 

•^ '  q  c       a 

2cosiC      1       1 

i. = I • 


coshA      cos^B      cosA(7      111 
p)  q  r  a 


b-^c 


(/3)    Again,  join  BD',  CD':  these  lines  will  be  equal  to  one 
another,  since  they  subtend  equal  angles  at  the  circumference. 
But  BD"  =  c'  +  p'  -  2pc  cos^^, 

GD'^  =  V'  +  p"  -  2pb  cos\A  ; 
.*.  c"  —  2p'e  cos^^  =  ¥  —  2p'b  cos^^ ; 
.•.  2p'  cos^-4  =  J  +  c. 
Similarly,  2q  cos  ^5  =  c  +  «, 

2r'  cos^(7  =  rt  +  i; 
.'.  p  cos^^  +  q  cos ^5  4  )•'  cos|  C  =  a  -\-  b  -{■  c. 


104  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

1851. 

1.  If  ABC  be  a  triangle  right-angled  at  (7,  E  the  point 
in  which  the  inscribed  circle  touches  BC^  and  F  the  point  in 
wliich  the  circle  drawni  to  touch  AB  and  the  sides  CL4,  CB 
produced  meets  CA :  shew  that  if  EF  be  joined,  the  triangle 
FEC  is  half  the  triangle  ABC. 

Let  r,  /  be  the  radii  of  the  inscribed  and  escribed  circles 
respectively ;  then,  since  0  is  a  right  angle, 
CE=r   and   CF=r\  and 
triangle  FEC  =  \rr' 

[s  —  a)  [s  —  h)  (s  — c)]*    {s[s  —  a)  (s  — Z>)]* 


=  i 


s 
adopting  the  usual  notation, 


=  i(6"-a)  {s-h) 


2 

=  ^  (Z>  +  c  —  a)  [a-^c  —  b) 

=  1  [a^  +  F  -{a-  h)']  -.'  c'  =  a'  +  b'\ 

=  ^ah 

=  I  the  triangle  ABC. 

2.    Shew   that    sin/3  sin  7  sin  [y  —  /3)  +  sin  7  sin  a  sin  (a  —  7) 
+  sin  a  slnyS  sin  {^  —  a.)  +  sin  (7  —  /3)  sin  (a  -  7)  sin  (/3  —  a)  =  0. 

We  have,  in  general, 

sin^  sin  5  slnC 

=  ^smA{cos{B-C)-cos[B+C)} 
=  i{sm{B+C-A)  +  sm{C+A-B)  +  sm{A  +  B-C) 
-mi{A  +  B+C)}. 
Hence,  if  ^  =  /3,  B=y,  C=y-B, 

sluyS  sin  7  sin  (7-/3)  =  i  (sin  2[y  —  /3)  +  sln2y8  -'feln27}. 

Similarly, 

sin7  sina  sin  (a  —  7)  =  4  {sin  2  (a  —  7)  +  sin27  —  sin2a}, 
sin  a  sin^  sin  (/3  —  a)  =  ^  {sin  2(/3  —  a)  +  sin  2a  —  sin  2/3}. 


1851.]  PLANE  TRIGONOMETRY.  105 

Again,  if  ^  =  7  -  ^,  i?  =  a  -  7,  C  =  /S  -  a, 
sin  (7  —  /3)  sin  (a  —  7)  sin  (yS  —  a) 

=  i{sin2  (iS  -  7)  +  sin2  (7  -  a)  +  8in2  (a  -  yS)} ; 
therefore,  adding  these  equations,  we  get 

sinyS  sin7  sin(7  —  yS)  +  sin7  sina  sin  (a  —  7)  +  sina  siii/9  sin  (/3  —  a) 
+  sin(7  — ;S)  sin  (a  — 7)  sin(/3  — a)  =  0. 

3.  The  equation  sina;  =  0  has  not  any  imaginaiy  roots. 

We  have  — ^  2  smx  =  e"""  —  e     '^. 

Now,  every  imaginary  quantity  may  be  expressed  under  the 
form  a  H — */3.     Substituting,  then,  this  quantity  for  a*,  we  get 

-Ij:  — ix  --,     -o  — i,     a 

£  —  £  =£"£/*  —  £      '    S.P 

=  cos  a  (e"''  -  £^)  +  -*  sina  {e~i^  +  z^) ; 
therefore,  if  sina;  =  0,  we  must  have 

cosa(£'/^-£/')  =  0, 
sina(£-/^  +  £'^)  =  0. 
These  require,  either  that 

cosa  =  0    and    £~^  +  e''  =  0, 
which  cannot  be  satisfied  by  any  real  vakie  of  /3 ;  or  that 
sina  =  0,   and  £~^  -  e''  =  0, 

which  can  only  be  satisfied  by  yS  =  0,  shewing  that  a;  =  a,  a  real 
quantity  :  whence  the  equation  sin  a;  =  0  has  not  any  imaginary 
roots. 

4.  If  the  cosines  of  the  angles  A^  B^  C\  of  a  plane  triangle 
be  in  arithmetical  progression,  shew  that  s  —  a,  s  —  h^  ^  ~  ^1  "^^ 
be  in  hannonic  progression,  s  being  the  semi-siun  of  the  sides. 

We  have 

cosvl  =  1-2  sin'^^,  cos5  =1-2  sin'^  B,  cos C  =  1  -  2  sin'^ (7; 

therefore,  if  cos^,  cos^,  cos 6*,  are  in  arithmetical  progression, 
sin'*^^,  sin'' ^5,  sin'^'^C,  are  so; 


lUG  .SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

^(^s-a){s-c)  _{s-a){s-b)       {s-c){s-b)^ 
ac  ab  CO 

2h  c  a 

+ 


'   s  —  h       s  — 


c       s  —  a 


or  2[^--(5-Z>)]  ^  s-[s-c)  ^  s-{s-a)  ^ 

s  —  b  s  —  c  s  —  a      '' 

2s  s  s 

+ 


s-b 


s  —  c      s  —  a 


s  —  b       s 


1  1 

+ 


c      s  —  a 


whence  ,  7  ,  ,  are  in  arithmetical  progression ; 

.-.  s  —  a  ,  s  -  b  ,  s  —  c,  are  in  hannonieal  progression. 


107 


SPHERICAL  TRIGONOMETRY. 

1848. 

1.    In  a  right-angled  spherical  triangle,  shew  that 
sin  a  tan^^  —  sinb  tan^J5  =  sin  (a  —  b) : 
shew  also  that  if  ^  be  the  spherical  excess, 

.    1  T-,      sinAasini&  ,  „      cosia  cos^b 

cos^c  cos^c 

(a).    We  have  in  general 

,     ,  1   —  C08-4 

tan*^  = ; — 5 —  , 

^  sin^      ' 

,     .      .       sin(7  .  sina 

and  sin>4  =  — —  sma  =  -; —  , 
suic  sine 

since  C  is  a  right  angle  ; 

.•.    sina  tan^^  =  sine  (1  —  cos^), 

=  8inc(l  —  tan6  cote)  by  Napier's  rules, 

=  sine  —  tanZ>  cose : 

similarly  sin 5  tan ^5  =  sine  —  tana  cose; 

.'.    sina  tan^^  —  sinJ  tan ^5  =  cose  (tana  — tanZ*), 

cose        .    ,       ,, 

=  7  sm  a  — 6  . 

cosa  cos 6 

But  by  Napier's  rules,     cose  =  cosa  cos^, 

.•.    sina  tan^^  —  sin  J  tan^^  =  sin  (a  —  b). 

(^).    Again, 
sin'^^a  sin^^i  +  cos^'^a  cos'^^J  =  :^{(1  —  C08a)(l  —  cosi) 

+  (1  +  cosa)(l  +  cosJ)}, 
=  ^(1  +co8a  cos  J), 
=  ^(1  +co8c)  by  Napier's  rules, 
=  cos"''^o; 


108  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1848. 

fsin^asin^bV      /cos^a  cos|/a'        •  217;.  .        2  if     /,\ 
I       -  ^_  \   _|.  / f — — ±_\   =  sin^i!/ +  cos^ii...(l) : 


V      cos^c      /        V      cos^c 

ami  in  any  spherical  triangle 

w^       r,\       cos^ia  —  b)        .^ 
^^  '       cos^(a4-J) 

also  .1  +  5  =  180°  +  ^  -  C,  hence 

cotMC-^)  =  ^i^|cot|C; 
^^  ^       cos(^a  +  ^>)        ^    ' 

therefore,  since  C  is  a  right  angle, 

1  +  tan^JS*      1  +  tan^a  tan  ^5 


1 

—  tan^^" 

1  —  tan^a  tan^J  ' 

.    tanij5;  = 

tan^^a  tan^J, 

sin^^ 

sin^a  'm\\b      cos|rt  cos^Z* 

cos^^ 

cos^^c         '         cos^c        ' 

), 

sin^^  = 

sin^a  sin^J 
cos^c       ' 

cos4^  = 

cos^a  cos^J 

therefore,  by  (1 


cos^c 
the  required  formnlse. 

2.  If  three  small  circles  be  inscribed  in  a  spherical  triangle, 
having  each  of  its  angles  120°,  so  that  each  touches  the  other 
two  as  well  as  two  sides  of  the  triangle,  prove  that  the  radius  of 
each  of  the  small  circles  =  30°,  and  that  the  centres  of  the  three 
circles  coincide  with  the  angular  points  of  the  polar  triangle. 

Let  ABC  (fig.  60)  be  the  triangle,  draw  the  great  circle  AD 
to  D  the  bisection  of  ^C;  let  0  be  the  centre  of  one  of  the 
circles,  through  0  draw  the  great  circle  EOF  perpendicular  to 
BG  and  intersecting  AD  in  E^  FE  will  be  a  quadrant ;  ajso  join 
^0  by  a  great  circle  BOH^  50^  will  bisect  the  angle  ABG\ 
draw  OG  the  great  circle  perpendicular  to  AD.  OG  and  OF 
Avill  each  be  equal  to  r  the  radius  of  the  small  circles :  let 
BG=2a,  BF=x,  FD  =  y. 

Then,  in  the  right-angled  triangle  ABD, 

coa  A  BD  =  tani5Z>  cot  ^5; 


1848.]  SPHERICAL   TRIGONOMETRY,  109 

or,  since  L  ABD  =  120°, 

1  1  —  tan'' a 

—  h  =  tana  — — , 

^  2  tana    ' 

1  —  tan''a 


2  ' 

.-.    tan'^a  =  2, 

and    cos  a  =  —^ . 

Again,  in  the  right-angled  triangle  OBF^ 
smBF=  cot  OBF.tan  OF, 

1 

or   smcc  =  — ^  tanr, 

and    in   the    right-angled    triangle    FO  G,    OF  =  90°  —  r,    and 
/  OFG  =  FD  =  2j,  and 

sin06^  =  sinO^sinO^^?, 
or   sin?'  =  cos?-  sin^, 
.•.  amy  =  tanr  ; 
and  cosa  =  cos(x  +  j/), 

=  (1  —  sin'''a;)*(l  —  sin^i/)*  —  sin.r  siny, 

or  =  (1  -  ^tan'''>-)4(l  -  tan'r)*  -  ^  tanV ; 

2 
.*.    cos'' a  -I-  rj  cosa  tan^'r  +  ^tan*r  =  1  —  ^tan^r  +  ^  tan*?-, 

or   i  +  I  tanV  =  1  -  |  tan^  r, 
.*.    tan'"'?-  =  ^, 
and  r  =  30°. 

Again,  let  BOH  =  h,  BO  =  z-    then,  in   the   right-angled 

triangle  BHCy 

cos  BC  =  cosBH.cosHC, 

or  cos  2a  =  cos  J  cosa, 

.*.  2  cos"''a  —  1  =  cosa  cosi, 

^^   -  3  =  51  C086,  •.•  cosa  =  -.  , 


and  cos5  =  —  wi) 


3*    '   •  -"^^^-34 
1 


110  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1849. 

also  in  the  right-angled  triangle  OBF^ 

sin  OF  =  sin  OBF.m\  OB, 

3* 
or    \  =  —  sin  z  : 
2        2  ' 

.'.   sin2;  =  — r  =  —  cosA, 
3*  ' 

.-.    J  =  90°  +  2;, 

and  0/f  is  a  quadrant. 

Again,  joining  OC,  we  have 

cosOC  =  cosO^  C0SJ5C  +  slnO^  sin  00  cos  OB C, 
=  cosz  cos2«  +  sinz  sin  2a  cos  60°, 
_  _  2^    1        1   2.2*  1 
~~3*'3'^P~3~2' 
=  0, 

and  00  is  a  quadrant,  so  is  also  0-ff;  therefore  0  is  the  pole 
of  -40:  similarly,  if  0',  0"  be  the  other  centres,  0'  is  the  pole 
of  CB,  and  0"  of  BA  ;  therefore  0  0'  0"  is  the  polar  triangle 
of  ABC. 

1849. 

1 .    If  P  be  the  perimeter  of  a  spherical  triangle,  of  which  the 
angles  are  A,  B,  0,  and  the  spherical  excess  F,  prove  that 

^.^^^^  [sin^E  smjA  -  ^E)  sin {B  -  ^E)  sin ( 0-  ^E)]^ 
^  2sin^^  sin^iJsin^O 

By  the  expression  for  the  sine  of  a  side  of  a  spherical  triangle 
in  temis  of  the  angles, 

{sini^sin(^-i^)sin(J5-i^)sin(0-i^)}i=isinasin5sinO, 
=  ^smh  sinO  sin^  =  |sinc  sin^  sinJ5, 

1  2 

=  ^(sina  sini  sinc)^(sin^  sin5  sinO)^ (1). 

Again, 

siniPslnfiP-a))i   .  (sin(^P-Z>)  sin  (^P-c)]* 

I         ^  [  sin 5  sine  J  ' 


COsi^  = 


^±   »un^2. 


(        smo  smc 

with  similar  expressions  for  the  cosines  and  sines  of  ^B  and  ^  0 ; 
cos"!^  cos^'iPcos'-'iO  sin^'iP 


sini--4  siniP  sinAO        sin  a  sin?;  sine  ' 


I  Hoi.]  SPHERICAL  TRIOONOMETRY,  111 

sinasinisinc  ~  sin'^^  sin''^i?sin''^C  ' 

_        sin''^  sin'^i?  sin'^6' 

.    1  r»      1  /  •        •    7    •     \3    (sin^  sinjB  sin (7) 
.".  sin^i^=  i(sma  sino  smc      .    ,    . — .    ,  „   . — rr^, 
^         *^  ^   sm^A  sm^B  sm^C^ 

_  {smEsmjA  -  ^E)  sm{B  -  ^E)  smjC  -  ^E)]i 

~  2sln^^  sin^Z?  sin^C  ' 

bj  (1);  the  required  formula. 

1851. 

1.  If  ABC  be  a  spherical  triangle,  right-angled  at  C,  and 
cosvl  =  (cosrt)^,  shew  that  b  +  c  =  ^tt  or  |7r,  according  as  b 
and  c  are  both  less  or  both  greater  than  ^tt. 

By  Napier's  rules 

cos^  =  cosrt  sin  5, 

but  by  the  conditions  of  the  problem 

cos^  =  cos'''a, 

.•.    cos  a  =  sin  5. 

Again,  by  Napier's  rules 

cose  =  cos  a  cos  J, 

=  sini?  cos  J  from  above, 

sinJ5   .    ,         , 

=    .    ,  smo  coso, 
smo  ' 

=  — —  smb  cos  J,  since  0  is  a  right  angle  ; 

.•.  sine  cose  =  sin?>  cosJ, 
or  sin2e  =  sin  26; 

and  b  is  not  equal  to  c,  as  then  B  would  be  a  right  angle,  and 
A  would  equal  «,  which  is  contrary  to  the  equation  cos^  =  cos^/; 
^euce  2&  +  2c  =  TT  or  Stt. 

Now  b  and  e  are  both  greater  or  both  less  than  ^tt,  since 
cos^  or  cos'^a  =  tan  J  cote ;  therefore  2J  +  2e  =  tt  or  37r,  according 
as  b  and  e  are  both  less  or  both  greater  than  ^tt. 


112 


THEORY  OF  EQUATIONS. 

1849. 

1.    Given  y  =  .rs",  prove  that 

-  =  1  +  717  +  ^  +  •••  +^T^'+  ••• 
X  [2         [3  \n 

One  root  of  the  equation 

?/  -  a-c^  =  0 
is  the  coefficient  of  -  m  the  expansion  of  —  log  [  1 )  . 

(See  Murphy's  Theory  of  Equations^  p.  77,  Art.  62,  and  p.  80, 
Ex.  3.) 

Now -log    1 =  — +-^  +  ...+ -+  ... 

V       y  I      y      ^  y  ^  y 

Expanding  the  exponentials,  we  see  that  the  coefficient  of  -  is 

^x'       i'x^  n"-'x 

2         1.2.3  1.2  .„  n  ' 

which  is  therefore  a  root  of  the  given  equation. 


\«-i 


-u-  y       ^       2x      i^xY  inx) 

Hence        ^  =  i  +  —  +  ^-—L  +  ...  +  v — L_  +  ... 
X  [2  [3  \n 

2.    If  a^j,  x^...x^  be  the  roots  of  an  algebraical  equation, 

and  no  two  of  them  be  equal,  then 

1  1  1  1 


jn-1 


7n  being  a  positive  integer  less  than  n. 


1849.]  THEORY   OF   EQUATIONS.  113 

(a).    Here    f{x)  =  {x-x;){x-x^),..{x-xj, 
we  may  therefore  assume 


1  A,  A  A 


f{x)       X  -  X^       X  -  x^ 


X  —  X 


1 


-4j,  A^y..A^^  being  independent  of  a;; 

.-.  l=A^{x-x;){x-x^)..,{x-xJ+A^{x-x;){x-x^)...{x-x^){x-'X^)+... 
+  A^^ {x  —  x^)...{x  —  a;,^_j)  identically ^ 

Hence,  putting  x  =  x^^ 

1  =  A^{x^-x^){x^-x;)...{x^-x„), 

=  Af{^.) (1); 

similarly    1  =  AJ'{x^) (2), 


i  =  A/'K) W- 

Hence 

1     ^  1  1_  _  1_  _ 

/ (^)  ~  (•» - a^i) />i)      (^ - a'J f\^,)      '"'^  ix- xj  f{^J 

identically. 
Therefore  putting  a;  =  0,  which  makes /(a?)  =^„, 

J_____l 1__  _  1 

\_  1  1_^  1 

_  1  /  ^         2CJ 


and  similarly  for  the  other  fractions.     Hence 
1     _       1       /        ^      ^  \      _1 


H-?  +  ^+-.l+z>,VJl+f  +  |{  +  ...) 


1        /,       rr        cr'           \ 
+  ...  +-rrr-s     1  +  --f  -^  +  ...     (1). 

1 


114  SOLl'TIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

B^t    J_  ^ 1 ^  1 

f{x)      x"  +2hx-'  +  ...  +2)„       ^n  /j  ^l\  _^       _^ A 

V        X      '"      x^ 

a;    V         •>c 
Hence,  if  m  be  a  positive  integer,  less  than  «,  the  coefficient 

of  ^  in  7^— r  =  0 ;  and  the  coefficient  in  the  right-hand  member 

X        J-[x) 

of  (1)  is 

1 —  J — ? i-        -I — « —  • 

^  ffl-l  ^  m-i  ^  m-i 

hence  J.    .  +  ^^tt — r  4-  • . .  +  ■^, — ^  =  0. 

1851. 

T,^  1  1      —     i^  « 

1.    If  x+ =  -  3;?, 

1  —  X  X  -^ 

then     (x  +  y-^  -  0)'-'  i^)    =  -  27  (^;  -  a>)  (^  -  a>y, 

where  (o  is  an  imaginary  cube  root  of  unity. 

1                   1  —  X 
Call  the  quantities  a;,    and ^ ,  ^/j,  j/^?  ^^d  3/3  5  then 

y^+y,  +  yz  =  -  ¥ (i), 

a^  1 

and  y,y,  +  yjj,^y^,  =  Yzr^-x~  ^^"^^ 

= 1 1  +  a;  -  1 

1  —  a;  X 

=  y,  +  y,■^y^-^ 

=  -3(i^  +  l) (2), 

also    y,y^y^  =  -  1 (3). 

Again, 

x^  1  (1  -  a;)" 

y^y^  +  y^yz  +  2/3V1  =  j-^r^  -  ^^fz^  +     ^ 

_       3x{\-x)  _ 

~       x{l-x)   ~         ^*^' 


1H51.]  THEOllY    OF    EQUATIONS,  11/ 

and    y'^y^  +  y^y^  +  y'^y^  +  y'^y^  +  y~y^  +  y;  y^^ 

=  -y^  {3  (p  +  1)  +  ^,3/3}  +  -^7(2), 
=  9p  (^;  +  1)  +  3,  by  (1)  and  (3) ; 

•••  y"y,  +  y"y.  +  y'y,  =  9i?  (i>  + 1)  +  e ...  by  (4)  and  (5). 

And  [y^  +  mj.^  +  a)'V3)'  =  j/^  +  y/  +  ^3"^  +  3^?/,y,  (?/,  +  6)yJ  4 

similar  tenns  +  ^y^y^^ 

{3/1  +  (*"  +  ^)  3/2}  +  similar  terms 
=  (-3i^r-  9(&)-l)  +  3(<y'''-l) 

{9p(^  +  l)  +  6}by(l),  (4),  and(5), 
=  -  27/  +  27/  (26)''  +  ft))  -  27^?  (2  +  w) 
+  27&)''',  since  a>  +  &)'•*  =  —  1, 
=  -  27  (/>  -  ft))  (p  -  ft)7'' 


12 


(   no   ) 


GEOMETllY  OF  TWO  DIMENSIONS. 

1848. 

1.  With  two  conjugate  diameters  of  an  ellipse  as  asymptotes 
a  pair  of  conjugate  hyperbolas  is  constructed ;  prove  that  if  one 
hyperbola  touch  the  ellipse  the  other  will  do  so  likewise ;  prove 
also  that  the  diameters  drawn  thi'ough  the  points  of  contact  are 
conjugate  to  each  other. 

Let  the  equation  to  the  ellipse,  referred  to  the  conjugate 
diameters,  be 

1^  +  ?-=' W- 

And  to  the  hyperbolas 

^1/  =  ^" (2), 

^I/  =  -  c' (3). 

(a)    In  order  that  (2)  may  touch  (1),  we  must  have 
^  _^  ,  ^ 

a  perfect  square,  in  which  case  we  shall  have  also 
X"      xy      y^ 
a         c         0 

a  perfect  square,  and  (3)  ^v^ll  also  touch  (1). 

(/3) .    If  the  above  expressions  be  perfect  squares,  we  see  that 
4    _  J_ 

-      ah 

•••  '  =Y' 

and   -  =  ±  -  ; 
X  a 


1848.]  GEOMETRY    OF   TWO    DIMENSIONS.  117 

and  \i  x'y  be  the  coordinates  of  the  pohit  where  (1)  meets  (2), 

,,       he'      W 

^    =2- 

Similarly,  if  x"i/"  be  the  coordinates  of  the  point,  where  (1) 
meets  (3), 

„„      b  ,,,,      a 

y'=2^  -^  =2- 

And  if  r\  r",  be  the  lengths  of  the  corresponding  semi- 
diameters,  CO  the  angle  between  the  axes, 

r"'  =  x'^  +  2^'^  —  2x1/'  cos  w  =  ^[d'  +  b^  —  2ab  cos  to), 
r'"'  =  x"^  +  y'"''  —  2x"y"  cos  g)  =  \[c^  +  i^  +  2ah  cos  oj), 
...  r"'  ^r"'  =  ce  +  h'; 
therefore  ?•',  r"  are  conjugate  to  each  other. 

2.  Shew  that  the  curve  which  trisects  the  arcs  of  all  seg- 
ments of  a  circle  described  upon  a  given  base  is  an  hyperbola 
whose  eccentricity  =  2. 

Let  AB  (fig.  61)  be  the  base,  a  its  length, 

AC  =  CD  =  DB  =  r,  CAB  =  0, 

we  then  have  r  {1  +  2  cos  6)  =  a, 

or,  referring  the  curve  to  A  as  origin  and  AB  as  axis  of  a;, 

x'  -\-  y'  =  [a-2xY'^ 

.-.  3j?'  -  /  -  4aa7  -|-  a'  =  0, 

the  equation  to  an  hyperbola,  the  squares  of  whose  axes  are  to 
one  another  in  the  ratio  3:1,  and  whose  eccentricity  therefore 

=  (3  +  1)^  =  2. 

3.  Let  Z)  be  a  point  in  the  axis-minor  of  an  ellipse  whose 

eccentricity  is  e,  S  the  focus,  0  the  centre  of  curvature  at  the 

D  9 
extremity  of  the  axis-minor ;  with  centre  D  and  radius  =  — — 


118  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

describe  a  circle  ;  shew  tliat  this  circle  will  touch  the  ellipse  or 
fall  entirely  without  it,  according  as  D  is  nearer  to  or  further 
from  the  centre  than  the  point  0. 

Let  C  be  the  centre  of  the  ellipse,  and  let  CD  =  h^  also  let 
a,  J,  be  the  semi-axes  of  the  ellipse,  its  equation  will  be 

a        o 
Also  DS'  =  d'^  +  li^^  so  that  the  equation  to  the  circle  will  be 

72 

a;"'  +  [y  +  ]if  =  a^  +  4  (taking  D  below  C), 

1  -  e' 
or  .-r^  4-  y  +  2/^ J/  =  «'  +  K'  — ^—  . 

Where  this  meets  the  ellipse,  we  have 

y    W      aV         «■'        o'      e'  "' 

or    ^-^/  -  2%  +  A''  -^  =  0; 

.-.3/  =  -^-  h. 

If  this  value  of  y  give  a  real  value  for  a-,  the  circle  will 
touch  the  ellipse,  if  not,  it  will  fall  entirely  without  it,  since  its 

radius  f  ct'  +  -5 )    is  greater  than  a,  and  therefore,  a  fortiori^  than 

DB\  which  is  less  than  h. 

In  order  that  the  value  of  x  may  be  real  it  is  necessary  that 
y  be  not  greater  than  Z>,  therefore 

\-e'  , 


e' 

or 

he' 
<-^-h, 

? 

<B0- 

BC\ 

<C0; 

1848.]  GEOMETRY    OF    T\V(»    DIMENSIONS.  119 

therefore  the  circle  will  touch  the  ellipse,  or  fall  entirely  without 
it,  according  as  h  <  or  >  CO^  i.e.  as  D  is  nearer  to  or  further 
from  the  centre  than  0. 

4.  PSp  is  any  focal  chord  of  an  ellipse,  A  the  extremity  of 
the  axis-major ;  AP^  Ap  meet  the  directrix  in  two  points  Q^  q : 
shew  that  I  QSq  is  a  right  angle. 

We  may  prove  this  property  for  any  point  in  the  ellipse 
by  a  process  exactly  similar  to  that  of  Part  I.  Conies^  1848,  3 ; 
except  that  we  have  the  equations 

sin  PBS  =  e  sin  PSR. sin  PRN^ 

sin  QR8  =  e  sin  QRS  sin  PRN, 

instead  of  those  there  given. 

This  theorem  may  also  be  proved  by  the  method  of  Reci- 
procal Polars.     (See  Salmon's  Conic  Sections^  chap.  XIV.) 

Take  the  polar  reciprocal  of  the  whole  system  with  regard  to 
the  focus  S.  To  the  ellipse  w^ill  correspond  a  circle,  to  the  point 
P,  2J^  two  parallel  tangents  Rt^  rt\  (fig.  62)  variable  in  position. 
To  A  (or  any  point  in  the  curve)  will  coiTespond  a  fixed  tangent 
tt\  and  to  the  directrix  the  centre  S'.  Hence  to  AP^  Aj)  will 
correspond  the  points  t^  t'  respectively,  and  to  Q^  q  the  lines 
St^  St'.  But  it  is  easy  to  see  that  the  lines  Sf,  St'  are  at  right 
angles  to  one  another ;  therefore  the  line  joining  the  points  Qj  j, 
subtends  a  right  angle  at  the  focus  S. 

5.  In  the  given  right  lines  AP^  AQ,  (fig.  63)  are  taken 
variable  points  ^j,  q,  such  that  Aj)  :  pP ::  Qq  :  qA  •  prove  that 
the  locus  of  the  point  of  intersection  of  Pq,  Qp  is  an  ellipse, 
which  touches  the  given  right  lines  in  the  points  P,  Q. 

Let  AP  =  o^  AQ  =  5,  Ap  =  a,  Aq  =  yS;  then  the  conditions 
of  the  problem  give 

a:a-oi::h-^:/3,     or  -  +  f  =  1 (1). 

an 


120  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

Take  AF^  AQ,  as  axes;  then  the  equations  to  Pq,pQ  respec- 
tively, are 

M  =  ' (^). 

M- (^)- 

o  a 

whence,  eliminating  a,  yS  from  (1),  we  get 

X  y 

0  a 

x^      xy      if      ^  X  y 


a 


ab       1/  a  h 


the  equation  to  the  locus  of  the  intersection  of  Pq^  p  Q^  which, 
since  the  square  of  half  the  coefficient  of  xi/  is  less  than  the  pro- 
duct of  the  coefficients  of  x^  and  3/^,  is  an  ellipse. 

When  £c  =  0,  we  have 

J,       J  ^  +  1  -  0, 

•'•    3/  =  ^ 

shewing  that  the  ellipse  touches  AP  in  P. 

From  considerations  of  symmetry  it  is  evident  that  it  also 
touches  AQ  m  Q. 

6.  A  parallelogram  is  constructed  by  di'awlng  tangents  at 
the  extremities  of  two  conjugate  diameters  of  an  ellipse ;  prove 
that  the  diagonals  of  the  parallelogram  form  a  second  system 
of  conjugate  diameters,  and  that  the  relation  between  the  two 
systems  is  reciprocal. 


1848.]  GEOMETRY   OF   TWO    DIMENSIONS.  121 

Let   the  equation  to  the  ellipse  referred  to   the  conjugate 
diameters  be 

<+C  =  i (.). 


b 

The  equations  to  the  tangents,  drawn  at  the  extremities  of 
these  diameters,  are 

a;  =  a,     x  =  —  a^ 

y  =  h,     y  =  -h] 

therefore  the  equations  to  the  diagonals  of  the  parallelogram 
thus  formed,  are  ^      ., 

M (^)> 

I=-f («)■ 

At  the  points  where  (2)  meets  (1),  we  have 
a  b 

therefore  the  equations  to  the  tangents  at  these  points  are 

M=±^' W' 

therefore  these  tangents  are  parallel  to  (3).  Hence  the  diagonals 
fonn  a  system  of  conjugate  diameters. 

Again,  the  equations  to  the  tangents  at  the  extremities  of  (3) 

M  =  ±^' («)' 

and  at  the  intersection  of  (4)  and  (5),  we  have  either 

cc  =  0,    or  3/  =  0, 

shewing  that  the  diagonals  of  the  parallelogram,  formed  by  the 
lines  (4)  and  (5),  are  the  first  system  of  conjugate  diameters; 
hence  the  relation  between  the  systems  is  reciprocal. 

7.  PSp  is  any  focal  chord  of  a  parabola  whose  vertex  is  A, 
prove  geometrically  that  AP^  A_p  will  meet  the  latus-rectum  in 
two  points  Q,  q,  whose  distances  from  the  focus  are  equal  to  the 
ordinates  of  the  points  j7  and  P  respectively. 


122  auLUTIUNS   OF   SENATE-HOUSE    PROBLEMS.  [1848. 

Draw  P3/,  prn^   (fig.  64)   ortllnates  to  the  points  P,  p  re- 
spectively.    Then  since  SQ  is  parallel  to  Pil/, 
.-.    SQ:AS::  MP :  AM, 
.-.   SQ:4.AS''::  MP:  ^AS.AM, 
::  MP:  MP'] 
.-.    SQ.MP=^AS\ 
Now   SP=2A8-\-  SM, 

=  2A8-\-  SPcoaPSM', 
.-.   SP{l-cosP8M)  =  2^^, 
.-.   Pil/(1  -  cosPSM)  =  2A8s{nP8M: 
similarly  pm[l  -\- cos 2)8 m)  =  2 A8  sinjy 8m  j 
.'.   PM.pm  =  4:A8'\ 

=  8Q.PM  from  above ; 
.-.  ^wi  =  8Q, 
similarly   PM  =  8q. 

Or  the  distances  of  Q,  q,  from  the  focus  are  equal  to  the 
ordinates  of  the  points  ^j,  P  respectively. 

8.  From  a  given  point  in  a  conic  section,  draw  geometrically 
two  chords  at  right  angles  to  each  other  which  shall  be  in  a 
given  ratio. 

The  construction  which  we  shall  give  depends  on  the  pro- 
perty that  all  chords  of  a  conic  section  which  subtend  a  right 
angle  at  a  given  point  P  of  the  curve,  intersect  the  normal  at  P 
in  a  fixed  point. 

Draw  PK  (fig.  65)  the  normal  at  P,  and  draw  PU^  PV  any 

two  chords  at  right  angles  to  one  another.     Join  UV,  cutting  the 

normal  in  K.   Then  by  the  property  above  enunciated,  if  PQ,  PR 

are    the    required    chords,    QR  will  pass  through  K.      Again, 

PR 
t3in  PQR  =  -^^  a  given  ratio,  hence  the  angle  PQR  is  known ; 

on  PK  we  describe  a  segment  of  a  circle  containing  an  angle 
equal  to  PQR ;  let  it  cut  the  ellipse  in  Q.  Join  PQ,  and  di'aw 
PR  at  right  angles  to  it,  PQ,  PR  will  be  the  required  chords. 


1848.]  GEOMETRY    OF   TWO    DIMENSIONS.  123 

9.  DeteiTnine  the  equation  to  the  conic  section  wlilch  passes 
through  five  points  whose  coordinates  are  given ;  and  thence 
shew  that  the  equation  to  the  conic  section  which  passes  through 
the  five  points  whose  coordinates  are 

1,-1;   2,  1;    -2,  3;    3,  2;   -1,-3, 
is  6iy  -  llxy  -  65a;^  +  36^/  +  174a;  -  151  =  0. 

Let  ajj,  y^ ;  x^,  y^ ;  a-,,  y^ ;  a„  y^ ;  a,,  y.^,  be  the  coordinates  of 
the  five  given  points,  which  we  shall  call  yl,,  A^^  A^j  A^j  A^  re- 
spectively. Then  the  conic  passing  thi-ough  A^y  A^^  A^j  A^^ 
circumscribes  the  quadrilateral,  whose  sides  are  A^A^j  -^3^4j 
A^A^^  A.A^.     The  equations  to  these  sides  are 

^-^2    y-y^  =0 


^2 

- 

^3 

X 

- 

^3 

^3 

- 

^4 

X 

- 

X, 

^4 

- 

^5 

X 

— 

^5 

y,  -  y. 

y-yz 
y,  -  y. 

=  0, 

y-y. 

=  0, 

y,-y. 

y  -y. 

=  0. 

^5  -^2    y,-  y^ 

Now  the  equation  to  a  conic,  circumscribing  a  quadrilateral, 
the  equations  to  whose  sides  are  ii^  =  0,  u^  =  0,  u^  =  0,  u^  =  0 
respectively,  is  ^  ,,  ^  -^^.^ 

2     4  3     0/ 

\  being  an  indetenninate  parameter. 

Hence  the  equation  to  a  conic  passing  through  A^,  A^^  A^j  A^^ 

'^  'X-  x^       y  -  y\  (X  -  x^       y-Jh} 


■^2  -  ^3    yt-  yJ  ' 

^^4  -  a^6    y,-  yJ 

fx-x^       y  -  yy 

^fx-x^       y-  y.^ 

v^3  -  ^4    3/3  -  y,^ 

'    \^S  -  ^2          3/5   -  3/2. 

=  \ 

The  quantity  X  is  detennined  by  the  condition  of  the  conic 
passing  through  the  point  A^{x^y^) ;  this  gives 

'^1  -  -^4    y,  -  y, 
^4  -  ^5    y*-  Vi 


'^1- 

3_ 

3/1- 

^ 

'^2- 

3/2- 

■yJ 

r^ 

-•^3 

_^ 

-y. 

y^  -  yj  v^6  -  ^2    .'/.-,  -  .'/■> 


124  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1848. 

Eliminating  X  between  tliese  last  two  equations  we  get, 
clearing  the  quantities  within  the  brackets  of  fractions, 

{[x-x^){y-y,)-[y-y,][x-x,)\\{x-x:^{if-y^)-{tf-y,)[x-x^)] 

{(^,-«'2)(3/2-^3)-(3^-2/J(^2-^8)}{(-'^-'^4)(3/-y5)-(yi-3^4)(^-a^6)l 

{ (-^ --^a) [y-yy  (3/ -3/3) (•^3--^4)  1  { (-^ --^J  iy-y-y  (i/-y.) i^^-^^)] ' 

the  equation  to  the  required  conic.     The  reduction  of  this  to  the 
symmetrical  form  would  be  very  tedious,  and  we  shall  therefore 
leave  it  in  the  above  shape. 
In  the  numerical  example 

^,  =  I7  3/1  =  -  1 ;   ^2  =  2,  3/,  =  1 ;   a-,  =  -  2,  ^3  =  3 ; 

^4  =  3,  2/4  =  2;    x^  =  -l,  y^  =  -3. 
Hence 

x^  —  x^  =  —  \  'y    a^j  —  cCg  =  3  ;        x^  —  x^  =  —  2  •    x^  —  x^  =  2j 

3/,-3/2  =  -2;  y,-y,  =  -4^;  y,-y,  =  -^]  3/^-3/5  =  2, 

a;^  —  X3  =  4  ;        x^  —  x^  =  —  b-,   x^  —  x^  =  A  ',   x^  —  x^  =  -  S^ 

?/.2  -  3/3  =  -  2 ;  3/3-3/4  =  1;      3/4-3/5  =  5;  3/5  -  3/2  =  -  4 ; 

therefore  the  above  equation  becomes 

{_  2(a;-2)  -  4(3/-!)}  {5{x-S)  -  ^y-2)] 
{(-2)(-l)-4(-2)H5(-2)-4(-3)} 

|l(^  +  2)  -  (-5)(y-3)}  {-i{x+l)  -  {-S){y  +  S)]  . 
{l(3)-(-5)(-4)}{-4(2)-(-3)2} 

(8  -  2a;  -  4j/)(-  7  +  5x-  4.y)  _  {- \Z  +  x -\- 6ij){b  -  Ix  +  3j/) 
•'•  20  ~  34  ' 

.-.    17(5a;-4?/-7)(a;  +  2j/-4)  -  5(a;  + 5?/- 13)(4a;-33/- 5) ; 

.-.    65a;'  +  llxy  -  61/  -  174a;  -  36y  +  151  =  0, 

or    61/  -  llxy  -  65a;'  +  36^/  +  174a;  -  151  =  0, 

is  the  equation  to  the  conic  passing   through   the  five  given 
points. 

10.  Two  chords  AB^  AC  are  drawn  from  a  given  point  A 
in  a  cm*ve  of  the  second  order  so  as  to  contain  a  given  angle, 
shew  that  BC  will  always  touch  a  curve  of  the  second  order. 


1H48.]  GEOMETRY   OF   TWO    DIMENSIONS.  125 

Let  the  equation  to  the  given  conic  section,  refen'ed  to  A 
as  origin,  be 

Ax^  +  2Bxi/  +  Cf  +  2l)x  +  2Ey  =  0 (1), 

and  let   ax  +  ^y  =  I (2) 

be  the  equation  to  -BC,  a,  /3  being  variable  parameters. 
At  the  points  of  intersection  of  (1)  and  (2),  we  have 
Ax^  +  2Bxy  +  Cy'  +  2  {Dx  +  Ey){ax  +  ^y)  =  0, 
or  ( C  +  2E^)y'  +  2  (5  +  ^a  +  D^]  xy^  +  (^  +  2Z>a)  x'  =  0...(3). 

This  may  be  considered  as  a  quadratic  in  "- ,  and  if  ^j,  ^2  ^®  ^^^ 

roots,  ij,  t,^  will  be  the  tangents  of  the  inclinations  to  the  axis 
of  X  of  AB^  A  C  respectively.  But  AB^  A  C  include  a  constant 
angle,  iarC^m  suppose;  hence  we  must  have 

-^ =  m. 

1  +  tA         ' 

or  ^  \,^' r^-^  =  irc. 

(I  +  M2) 


Now  from  (3),  by  the  the  theory  of  equations 


B+Ea^D^  _A  +  2l)a 

C+  2EI3      '      ''      C  +  2^/3 ' 

.-.  4i{{B+Ea  +  I)^Y-{A  +  2l)a){C+2E^)] 

=  m'  { {A  +  2Da)  +  C  +  2E^]\ 

.'.  4.[E'  -  AC-\-2{BE-CD)  a  +  2{BD-AE)  yS  +  E^a' 

+  D'0'  +  2{B+DE)a^], 

=  m'{A  +  C  +  2{I)a  +  E^)]% 

which  may  be  written  under  the  foi-m 

aa'  +  2hafi  +  c^'  +  2(f7a  +  e/9)  +  1  =  0 (4), 

a,  5,  c,  <7,  e  being  certain  determinate  functions  of  A^  B^  (7,  Z>,  J?, 
and  in. 

Now  consider  the  conic  section  whose  equation  is 

^V  +  2B'xy  +  Cy  +  2[D'x  +  E'y)  +  1  =  0 (.5). 


126  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

Where  (2)  meets  this,  we  have 
^V  +  Wxij  +  Cy  +  2{D'x  +  E'7/){aa:  +  ^i/)  +  [ax  +  ^i/Y  =  0; 
.-.  {A  +  2D' a  +  a')  x'  +  2  (5'  +  E'a  +  D'^  +  ayS)  xy 

+  {C'-{-2E'^  +  ^')f  =  0. 

In  order  that  (2)  may  touch  (5)  the  roots  of  this  equation, 

considered  as  a  quadi'atic  in  - ,  must  be  equal ;  we  must  therefore 
have 

[B'  +  E'a  +  Z>'/S  +  a^Y  =  [A'  +  2D' a  +  a')  ( C  +  2E'^  +  /3'-') ; 
...  B"  -  A'C  +  2  {B'E'  -  CD')  a  +  2  [B'D'  -  A'E)  yS 

+  {E"'-C')  oi'  +  2{B'-  D'E)  a/3  +  [D"  -  A')  ^'  =  0, 
which  agrees  with  (4)  if 
E"'  -  C'_  _         B'  -  D'E'  _  D"  -  A'    _ 


B^  -A'C'~    '    B"  -A'G'~   '    B''  -  A  C 

B'E'  -  CD'  _        B'D'  -  A'E'  _ 
B"'-A'C   ~    '      B'-'-A'C   ~^' 

which  five  conditions  can  be  satisfied  by  means  of  the  five 
disposable  quantities  A'^  B',  C,  D',  E .  Hence  BC  always 
touches  the  conic  whose  equation  is  (5). 

This  theorem  may  also  be  proved  by  the  method  of  reciprocal 
polars.  For  taking  the  polar  reciprocal  of  the  whole  system 
with  regard  to  ^ ;  to  the  conic  will  coiTespond  a  parabola,  and 
to  -S,  C  (two  points  the  line  joining  which  subtends  a  constant 
angle  at  the  origin)  will  correspond  two  tangents  containing 
a  constant  angle.     The  reciprocal  theorem  then  is : 

If  two  tangents  be  dra\\Ti  to  a  parabola,  including  a  constant 
angle,  the  locus  of  their  point  of  intersection  is  a  curve  of  the 
second  order.* 

This  may  be  proved  as  follows : 

*  Taking  the  polar  reciprocal  of  this  systena  mth  regard  to  the  focus  of  the 
parabola,  the  theorem  to  be  proved  is  the  following : 

If  a  chord  of  a  circle  subtend  a  constant  angle  at  a  given  point  of  the  curve, 
it  always  touches  a  circle,  which  is  knoAvn  to  be  true. 


1848.]  GEOMETRY    OF   TWO    DIMENSIONS.  127 

Let  y  =  tx  +  - 

be  the  equation  to  any  tangent  to  a  parabola.     This  may  be 

written  ,,         „ 

t--y.t  +  -^0 (1). 

XX 

This   equation,   considered   as  a  quadratic   in    t^   gives  the 

tangents  of  the  inclinations  to  the  axis  of  the  two   tangents 

di'awn  to  a  parabola  through  a  point  {xy).     In  order  that  these 

may  include  a  given  angle  tan~S?i,  we  must  have,  if  ij,  ^^  be  the 

roots  of  (1),  t  —  t 

-^ =  m  ; 

1  +  f,t, 

therefore,  by  the  theory  of  equations. 


1  +  ^ 

X 


or   y^  —  4acc  =  nf[a  +  x)'\ 

the  equation  to  the  locus  of  xy^  which  is  therefore  a  curve  of  the 
second  order. 

11.  Pf  J)  are  the  extremities  of  two  semi-conjugate  diameters 
of  an  ellipse  E^  whose  semi-axes  are  «,  b ;  upon  FD  describe 
an  equilateral  triangle  PDR^  so  that  the  point  R  may  fall  with- 
out the  ellipse ;  the  locus  of  R  will  be  an  ellipse  E^ :  assuming 
the  above  result,  shew  that  if  E^  be  similarly  treated,  as  also  all 
the  successive  ellipses,  the  axes  A^^  B^  of  the  a;'^  ellipse  E^  so 
described  will  be  comprised  in  the  fonnula 

[a  +  5)(coti7r)-'  ±  (a  -  J)(cot^7r)*". 

In  the  figm-e  (66),  let  CP^  CD  represent  the  equal  semi- 
conjugate  diameters  of  the  ellipse  E^  and  let  D'  be  the  other 
extremity  of  the  diameter  through  D.  Join  PD^  PD'\  on  them 
describe  the  equilateral  triangles  PDR^  PD'R  ;  then  i?,  R'  will  be 
the  extremities  of  the  axes  of  E^.  Join  CR^  CR'.  Then 
CR  =  ^A^,      CR  =  \B^: 


128  SOLUTIONS   OF  SENATE-HOUSE  PROBLEMS.  [1848. 

also  ^,  =  t&nPCR  =  -,  and  FV'.CV  =  ^ab; 
CV  a'  ^      ' 

.-.   OF'  =  i«,   and  PF'  =  ij. 

And    CR'  =  CV  +  V'B', 

=  _  +  2iZ>cos-. 

O  TT 

Similarly  Ci?  =  ^  +  2*a  cos  -  , 
2*  6 

l(^,  +  JJ  =  («  +  &)(2*cos^  +  i), 

AT  ^X  7^  1  3^+1  2* 

Now    2*cos-+-,  =  -^^  =  3P33 

3^+l\|       /2  +  3hi       /l  +  cosi7r\| 


3^-1/        V2-3V        Vl-cos^TT/' 
=  lcot^)*. 


Similarly  it  may  be  shewn  that 


2^cos|-i  =  (tan^,   , 


.-.    i(5,  +  A)  =  («  +  ^)(cot^), 

i(5,-^J  =  (a-J)(tan^y, 
a  formula  connecting  the  axes  of  any  two  successive  ellipses  ; 

.-.   ^^  =  («  +  5)(cot^)    4-  («_J)(^tan^)  , 

A  =  («  +  ^)  (cot  ^)    -  (a  -  &)  (tan  ^)  . 


1849.]  GEOMETRY    OF   TWO    DIMENSIONS.  129 

Now  assume 

B^  =  {a  +  b)  (cot  ^)*'  -  -^  («  -  b)  (tan  ^)'", 

A=(«  +  ^)(cot^)     +-'(«-^)(tan^)   ; 
tlieu,  by  the  formula  already  given, 

^^^  =  -—T-"  (cot  -  )   +  ^.r-^  (tan-)  , 


2  V        127  2  V        127 

=  («  +  J)(cot-j        +_-(a-J)(tan^j        ; 

•Ml       A  A   +  B    /        7r\*-      A   -  B    f        ir\^ 

similarly   J,,,  =  ^-^  (^cot  -  j   +  ---^^  (^tan  -j  , 

=  («  +  J)(cot^)        +_-(«- J)  (tan  ^j        . 

If  then  the  assumed  fonn  hold  for  E^  it  is  proved  to  hold  for 
^^j.  But  it  has  been  shewn  to  hold  for  E^^  therefore  it  holds 
universally. 

1849. 

A  is  the  origm  (fig.  67),  5  a  point  in  the  axis  of?/,  BQ  a  line 
parallel  to  the  axis  of  ar ;  in  AQ  (produced  if  necessary)  P  is 
taken  such  that  its  ordmate  is  equal  to  ^^:  shew  that  the  locus 
of  P  is  a  parabola. 

Let  AB  =  a,  AP  =  r,  BAP  =  ^ir  -  6^  then  the  ordinate 
of  P=  rsin^,  also  BQ  =  «cot^; 

.'.    r  sin^  =  acot^, 
or   r^  siv^  6  =  ar  cos  6 ; 
therefore,  putting  r  sin  0  =  y^  r  cos^  =  a-, 

y'  =  c^, 

shewing  that  the  locus  of  P  is  a  parabola, 

a®      b^ 
2.    K  from  points  of  the  curve  — ^  +  -a  =  (a*  —  b^j  tangents 

be  drawn  to  the  ellipse  -7  +  75  =  1,  the  chords  of  contact  will 
'^       a       b 

be  normal  to  the  ellipse. 

K 


130  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 


If 

at  the 

1,  7]  be  the 
point  (.r^), 

current  coordinates  of  the  normal 
its  equation  will  be 

to 

the 

ellipse 

X 

a' 

v-y 
"    y   ' 

01 

r  - 

a^       1, 

T2                    2                          '  • 

a"  —  h^  X      1/  —  (^  y 
If  ar^,  y^  be  the  coordinates  of  the  pole  of  this  line  with 
respect  to  the  given  ellipse,  we  have 

_       g"       1 

-      ^'       1 
y^  -  y^  -a^  y- 

Eliminating  jc,  y  between  these  two  equations  and  the 
equation  to  the  ellipse,  we  get 

?  +  p  =  («■'-*')■'' 

the  equation  to  the  locus  of  the  pole  of  the  normals  to  the 
ellipse. 

Hence  if  from  points  of  this  curve,  tangents  be  di'awn  to  the 
given  ellipse,  the  chords  of  contact  will  be  normal  to  the  ellipse. 

3.  An  oblique  cone  stands  on  a  circular  base;  prove  that 
one  of  the  axes  of  the  section  made  by  a  plane  passing  through 
the  centre  of  the  base  and  pei-pendicular  to  the  axis,  is  a  mean 
proportional  between  the  other  axis  and  i)seca,  where  D  is  the 
diameter  of  the  base,  and  a  the  angle  between  the  axis  and  a 
normal  to  the  base. 

Let  MPM'  (fig.  68)  be  the  section  through  0  the  centre  of 
the  base  and  perpendicular  to  CO  the  axis  of  the  cone ;  let  the 
section  intersect  the  circular  section  RPR'  in  the  line  NP. 

Let  a  and  h  be  the  axes  of  the  section,  then 

F  _      NP'      _  RN.NR' 

a'  ~  MN.NM'  ~  MN.NM' ' 


1849.]  GEOMETRY   OF   TWO    DIMENSIONS.  131 

Let  lCAO  =  ^,   CBO  =  y;  also  the  z  MO  A  is  given  equal 
to  a.     Hence 

BN  _  sin(/3  +  a)        NR'  _  sin(/3'-a) , 
MN~      sin/3      '     :?^~      sm  13'      ' 

Z)"'' _  sin (/3  +  a)    sin(y8'  — a)  /-^ 

''  ^~       sin/3      •       siu^Q^       ^  ^' 

and  a  =  JL¥'  =  .1/0+  OJ/'=^l-.-^S^  +  _.J^l  ..(2). 

2(sm(/3  +  a)      sm(/3'-a)] 

Again,  since  AO  =  OB, 

sm  AC  0      sinBCO 
sinCAO  ~  smOBC 

cos  (8  +  a)      cos  f/S'  —  a) 

or  ^; — pj-^  =  — \ — 7^7 — ^  =  p  suppose ; 

snip  smp  -^      ^^ 

^      ;:>  +  sina  ,^,      ^  -  sina 

.•.  cGt/3=^^— — -,     cota  =~ . 

cos  a  cos  a 

Substituting  in  equations  (1)  and  (2),  we  have 

12 

-jj  =  (cosa  +  cotyS  sin  a)  (cos  a  —  cot/3'  sina), 
(1  +2^  sina)(l  —])  sin'a). 


cos  a 

_  1  — ^^sin"''a^ 
cos'^  a      ' 

,  D  I      cosa  cosa 

and  a  =  — -. y t     , . 

2   VI  +p  sin  a       1  —  ^?sina/ 

I)  cosa 


1  —  p  sm  a 

.-.   =  jr,  i>seca, 

or   h''  =  aD  sec  a, 
and  5  is  a  mean  proportional  between  a  and  Z^seca. 

4.  Let  P,,  P^,  Pj,,  (),,  (2.,?  ^3  be  six  points  lying  in  a  conic 
section;  let  the  areas  of  the  triangles  P^Cl^Q^,  ^iQaQ^i  ^iQiQ^i 
be  denoted  by  ^,,  P,,  O,,  and  the  areas  of  the  triangles  formed 

K2 


132  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

by  putting  /!,,  P^  successively  in  the  place  of  P,  be  denoted  by 
^'11  ^27  ^2)  -^3)  A)  ^3  I'espectlvcly  ;  then  will 


A,\B,C,     B,CJ  ■  A^\B,C\      BfiJ  '  A,\B,G,      B,C„ 

Let  Mj,  Mj,  Wg  denote  the  distances  of  any  point  from  the  lines 
^•2^35  ^3^15  QiQ;^  respectively,  then  ?/,  =  0,  tt,^  =  0,  u^  =  0  will 
be  the  equations  to  these  lines  themselves.  And  the  equation  to 
any  conic  section  passing  through  Q^^  Q^^  Q^^  may  be  written 
under  the  form 

^  +  ^^  +  -^  =  0 (1), 

W,  M,  «3 

Aj,  X,^?  \  being  constants  whose  values  will  be  determined  by 
the  condition  of  the  conic  passing  through  P^,  P^,  Pg. 

Wg',    ^3'  be  the  values  of  u^^  w.^,  u^  respectively  at  Pj, 


Let  u 

u. 
u 

Then   A  =  4^.^3-<,   A  =  ^^3^x-<,    ^,  =  i^x^.-<, 
with  similar  expressions  for  ^.^,  P^)  ^2  5  -^35  ^35  ^3  • 


J  <'?  <   ^j 

1     5<"5<"    ^- 


1    \        1 

\BA 

- 

i^.cj  + 

1 

A 

U<^2 

1 

8 

1'   f 

1 

1  \ 

"     ^2^3-a^, 

'Q,qA<\^. 

^3 

^2 

"<'i 

1  /    1 

%             ^ 

1 

7J^.," 

Ct 

1 

>3" 

«2«*2/i 

3  ^ 

And  since 

P.P^Pg  all  lie 

in  (1), 

we 

have 

=  0, 

A/,  A.„  A„  _ 

-V    +   -T,     +  -7,  =0, 
%,             M,             M3 

A..  A-„  A<-  ^ 

-7^  +  -^,  +  -^  =  0 ; 

i*.  W.,  t*„ 


1849.]  GEOMETRY   OF   TWO    DIMENSIONS.  133 

whence  eliminating  "^^W  by  cross-multiplication,  we  get 

1/1  1     \        1    /     1  1 

+ 


1/1  1  , 

+  — ,     -7^  -  -^r-.    =0; 

whence  dividing  by  ^Q.,Q^.Q^Q,.Q,Q^, 
1/1  1\        1/1  1\        1/1  1, 


5.    The  equations  of  three  straight  lines  are 

u  (=  X  sin  6  —  1/  cos  ^  +  c)  =  0,     «,  =  0,     m^^  =  0 ; 

prove  that  the  equations  of  the  four  circles,  to  each  of  which 
these  lines  are  tangents,  are 

.   .     0-$^         .    .     0-  0^  0       0 

w*  sm  -^— — i  +  ?/j*  sm   — - — ^  +  u^  sm  -i— —  =  0, 

a  a  i 

w*  sm  -^-- — -  +  w  *  cos  ?  +  ?/ *  cos  — =  0, 

2  2  2 

*  ^.  -  ^1  I    '     Q-Q.  *  ^,  -  ^ 

w*  cos  -^^—r — i  +  w  *  sm  — - — ^  +  ^C  cos  -^ =  0, 

2  2  2 

*  ^2  -  ^1   .       i         ^  -  ^.          *    •     ^,  -  ^       . 
M*  cos  -^-— — J  +  u^  COS  — - — ^  H-  ?<2*  sm  -^ =  0. 

^  ^  ^ 

The  equation  to  any  conic  section,  touched  by  these  three 
lines,  may  be  wi-Itten, 

\u^  +  \u^  +  \u}  =  0 (1). 

K  we  reduce  this  equation  to  the  form  of  an  equation  of  the 
second  degree,  the  coefficient  of  the  terms  of  two  dimensions 
will  not  contain  c,  c^,  c.^;  hence,  to  find  the  condition  of  this  repre- 
sentmg  a  circle,  suppose  c,  c^,  c^  to  be  Indefinitely  diminished,  the 
ratios  c  :  c^  :  c^  remaining  unaltered. 

The  three  lines  u  =  0,  u^  =  0,  ?/,^  =  0,  will  then  all  pass 
through  the  origin,  and  the  circle  touching  them  will  degenerate 
into  the  origin ;  Its  equation  will  therefore  be 

x^  +  if  =  0, 
or   1/  =  ±  — *  .r. 


134  s()h:ti()ns  of  senate-house  problems.         [1849. 

Equation  (1)  will  therefore  become,  dividlug  out  by  a;, 

\  (sin  d  +  -i  cos  (9)4  +  \  (sin  6^  +  -*  cos  ^J* 

+  \  (sin  e^  +  -4  cos  ^Ji  =  0 (2)* 

Hence,  by  Demoivre's  theorem,  equating  real  and  imaginaiy 

parts  separately  to  zero, 

ff  0  ff 

X  sin  -  +  \  sin  -^  +  \  sin  ^  =  0, 

2  2  2  (  ^^y 

e  e  0 

\  cos  -  +  X,  COS  -~  +  \,  cos  -^  =  0. 

2  2         -^         2 

Eliminating  X,  X^,  X,^  by  cross-multiplication  from  (1),  (3), 
Ave  get 

1       •  ^.,    ~     ^,  i.        •  G    ~    ^.,  i        •  ^1     ~    ^  /x 

w*  sm  -^'— : — -  +  w,^  sm  — - — -  +  uj  sm  -^— —  =  0. 
2  '  2  ^  2 

Now  if  in  this  equation  we  write  tt  +  ^  for  ^,  tt  +  ^^  for  6^, 
and  rr  +  6,^  for  0^,  successively,  by  which  substitution  equation 
(2)  is  not  altered,  we  get  the  equations  to  the  remainmg  circles : 
these  are 

X    •     0.^  —  ^,  1         0  —  0„  1         0,-0      ^ 

tt*  sm  -~- — ^  +  M  *  cos  — - — ^  +  uj  cos      ^       —  0, 
2  2  2 


u^  cos 


0-0^         ^    .     6'  -  ^,         ,         ^  -  ^ 


2 


-^  4-  M  4  sin  — - — -  +  u}  cos  -^ —  =  0, 
"^  2  2 


,  0   —  0  1  0  —  0  i     '     0   —  0 

w*  cos  -^— — -  +  Wj*  cos  ^  +  Mjj*  sm  -^-— —  =  0, 

ia  i^  ^ 

the  required  equations  to  the  circles. 

1850. 

1.   If  at  a  given  point  two  circles  intersect  and  their  centres 
lie  upon  two  lines  at  right  angles  to  each  other  through  the 


*  This  method  of  investigating  the  condition  that  equation  (1)  may  repre- 
sent a  circle,  is  due  to  Mr.  Leslie  Ellis.  It  may  be  shewn  in  precisely  the 
same  manner,  that  if  ip  (m,  u^,  u^)  =  0  be  any  equation  of  the  second  degree,  the 
condition  that  this  may  represent  a  circle  is  </>  (^"^e,  £-i9i,  t~i6j^—o. 


1850.]  GEOMETRY    OF    TWO    DIMENSIONS.  135 

point;  prove  that,  whatever  be  the  magnitude  of  the  circles, 
their  common  tangents  will  always  meet  in  one  of  two  straight 
lines  which  pass  through  the  given  point. 

Take  the  given  point  as  origin,  and  the  lines  on  which  the 
centres  lie  as  axes.  Let  a,  )8,  be  the  radii  of  the  circles.  Then 
the  intersection  of  the  common  tangents  must  always  lie  on  the 
line  joining  the  centres  of  the  circles,  whose  equation  is 

?  +  ^=l. 

From  considerations  of  symmetry  it  is  easy  to  see,  that  if  this 
intersection  always  lie  on  one  of  two  fixed  lines  passing  through 
the  origin,  the  equations  to  these  lines  must  be  x  +  y  =  0, 
X  —  y  =  0.  Hence,  if  such  be  the  case,  the  equation  to  one  of 
the  common  tangents  must  be 

X       y      X  +  y 

-  4-  I  +  ^  -1=0, 

a      ^  7 

where  7  is  a  constant  to  be  determined. 

In  order  that  this  line  may  touch  the  circle  whose  radius  is  a, 
it  is  necessary  and  sufficient  that  its  distance  from  the  centre  of 
the  circle,  whose  coordinates  are  a,  0,  be  a.  We  must  there- 
fore have 

1    i\'    /I    1 Y"      ' 

a      7/       Vy9      7/ 

.}__(]_      ly      (I       1 

'  ■  7'  ~  Va  7/  V/3  7 
shewing  that  7  is  a  symmetrical  function  of  a  and  yS,  and  there- 
fore that  if  this  straight  line  touch  one  of  the  circles,  it  must  also 
touch  the  other.  Hence  the  intersection  of  the  common  tangents 
always  lies  on  the  line  a;  +  y  =  0,  if  a  and  yS  have  the  same 
sign,  i.e.  if  both  centres  lie  on  the  positive,  or  both  on  the  nega- 
tive side  of  the  origin.  If  one  centre  lie  on  the  positive,  the 
other  on  the  negative  side,  similar  reasoning  will  shew  that  the 
intersection  of  the  common  tangents  lies  on  the  line  x  —  ?/  =  0. 


136  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1850. 

2.  A  number  n  of  equal  confocal  parabolas  are  ranged  all 
round  the  focus  at  equal  angular  intervals ;  shew  that  the  product 
of  the  distances  of  all  the  points  of  intersection  from  the  focus 

is  — 5—  ,  I  beinff  the  latus-rectum. 

Taking  the  common  focus  as  pole,  the  equations  to  the  para- 
bolas will  be 

'  =  -b ("' 

4  sm    - 
2 

-—77B~^ <^'' 

V2  ^  n) 

*  ™  (2  +  it) 


4  sm     -  H TT 

V2  n 

If  ^j,  0^  be  the  two  values  of  6  at  the  intersections  of  (1)  and 

{m  +  1)  it  is  manifest  that  since  these  intersections  lie  at  the 

extremities  of  the  same  chord  passing  through  the  pole, 

Also  we  easily  see  that     u,  = , 

.'.    C/    =  TT . 

^  n 

Hence,  if  r^,  r^  be  the  corresponding  values  of  ?•, 
/  I 

'  .  ,  mTT '       ■■^       ,        .,  mTT ' 

4  sm   -—  4  cos   — - 

271  2n 

•  •.    rr   =  —  . 
^  ^  .   „  mTT 

4  sm   — 
n 


1850.]  GEOMETRY    OF   TWO   DIMENSIONS.  137 

Similar  expressions  holding  for  the  intersection  of  (1)  with 
each  of  the  other  parabolas,  we  have 

product  of  all  the  distances  of  intersections  of  (1)  with  the  other 
parabolas 


^     '  Sin  —  sni    — sm  tt 

w  n  n 

-r,  ,  .     TT    .     27r  .     n  —  \  n 

rJut  sm  -  Sin  —  ...  sin  tt  =  -^,  ; 

n  n  n  2 

^2(«-l) 

therefore  the  above  product     =  - — 5—  . 

n 

To  get  the  product  of  all  the  distances  of  intersections,  we 

n 
have  merely  to  raise  this  quantity  to  the  power  - :  (not  n  ;  since 

each  intersection  would  then  be  counted  twice  over) ; 
therefore  product  of  all  the  distances  of  intersections  =  — s-  • 

3.  The  locus  of  the  points  from  which  a  circle  is  projected 
into  a  circle,  upon  a  plane  inclined  at  a  finite  angle  to  that 
of  the  given  circle,  is  an  equilateral  hyperbola. 

Let  0  (fig.  69)  be  the  centre  of  the  given  circle,  AB  that 
diameter  of  it  in  which  it  is  cut  by  a  plane  through  0,  pei-pen- 
dicular  to  the  line  of  intersection  of  the  plane  of  the  circle,  and 
the  plane  of  projection.  Let  CD  be  the  corresponding  diameter 
of  the  circle  in  which  it  is  projected.  Join  CA^  DB^  and 
produce  them  to  meet  in  E^  E  will  be  the  point  from  which 
the  given  circle  is  projected. 

Draw  EG  parallel  to  CD  and  equal  to  AB^  terminated  by 
EC^  ED :  let  EG,  AB  intersect  in  P,  then  must  AP  =  GP^ 
BP  =  FP.  Through  0  draw  two  lines  OX,  0  F,  parallel  to 
those  respectively  bisecting  the  angles  APE,  APG,  and  take 
them  as  axes.  Let  «,  h  be  the  coordinates  of  P;  —  «,  —  5  of  ^  ; 
X,  y  those  of  E.  Then  x  will  be  the  abscissa  of  P,  and  it  is 
hence  casv  to  sec  that  n  -\-  2.r,  —  h  will  be  the  coordinates  of  G. 


138  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1850. 

Hence,  f ,  77  being  cuiTent  coordinates,  the  equation  of  ED  will  be 
f  —  a  7}  —  h 


a  +  2x  —  a      —  h  —  b^ 

^  —  a          7}  —  b 
or  -= —  — 

(1) 

X                  b 

V    / 

And  tlie  equation  to  EP  is       ^  =  x 

(2) 

At  E^  the  intersection  of  these  two  lines,  we  have  f}  =  V-,  and 
from  equations  (1),  (2), 

x~b' 
.'.  xy  =  aJ, 

shewing  that  the  locus  of  E  is  an  equilateral  hyperbola,  of  which 
OX^  0  Y  are  the  asymptotes. 

4.    Prove    that    y  =  nx  -\ ad  iniinitum   is   the 

^  X  +  X  +  ...  '' 

equation  of  a  hyperbola.  Find  the  position  and  magnitude  of 
the  axes,  and  write  down  the  equation  of  the  conjugate  hyper- 
bola under  the  same  form. 

Since  y  =  7ix  + 


y  —  nx  = 


X  -{-  X  -\- 

1       1 


X  +  X  +... 

1 

X  -Y  y  —  nx^ 

-  1 

y  —  [n—l]  x^ 

•••    {^J-nx){y-{7i-l)x]  =  1 (1), 

the  equation  to  an  hyperbola,  whose  asymptotes  are  represented 

by  the  equations  ,        ,  ^ 

"^  ^  y  =  nx^     y  =  (7?  —  1)  x. 

The  axes  bisect  the  angles  between  the  asymptotes,  therefore 


l.SaO.]  GEUMETliY    UF   TWO    DIMENSIONS.  139 

their  equations  are 

y —  nx    _  y  —  [n—\)x  , 

{\  +  ny  "  {1  +  {n-\y\^ ^^^' 

y  —  nx  y  —  i^n—\)x 

(Tm??  "  "  {i  +  (/i-ir}i ^^^ 

respectively.     Hence  the  positions  of  the  axes  are  known. 

To  find  the  magnitudes  of  the  axes,  we  have,  combining  (1)  (2), 

y  —  nx  = , 

{l  +  (n-iyf 

y  -  [n-l)x  =  ^ 5^ ^  ; 

(i+«'r 
.      {i+(^-iri^-(i+^-o^ 

•  •  »^  —        '  • 

_n{l  +  (n-l)']^-  (n-l)(l  +  ny 
{l-flw-irni+n")' 

=  2  [(1  +  71^)4  [1  +  {n-  lY]i  -  [n'  -  n  +  1)], 

which  gives  the  square  of  the  magnitude  of  one  of  the  semi-axes. 
Similarly  the  square  of  the  other  semi-axis  may  be  shewn  to  be 

=  2  [(1  +  riy  {1  +  (n  -  1)^}4  +  {n'  -n  +  1)]. 

The  equation  to  the  conjugate  hyperbola  is 

{y-nx)[y-{n-\)x]  =-1, 

which  may  be  written 

1 

11  —  nx  — , 

^  X  ■\-  y  —  nx^ 

_         1        1 


y 


X  —   X  — 

1       1 


.r  —  X  — .. 
is  the  equation  to  the  conjugate  hyperbola. 


140  SOLUTIONS  OP  SENATE-HOUSE   PROBLEMS.  [1850. 

5.  From  a  point  0  (fig.  70)  are  drawn  two  lines  to  touch 
a  parabola  in  the  points  P  and  Q\  another  line  touches  the 
parabola  in  R  and  intersects  OP,  OQ  in  8^  T;  if  V  be  the 
intersection  of  the  lines  joining  PT^  QS  crosswise,  0,  i?,  V  are 
in  the  same  straight  line. 

Let  OP  =  a,  OQ  =  h,  then  the  equation  of  the  parabola 
referred  to  these  lines  as  axes,  is 

!)'  +  (!/- (')• 

Let  08=  oi,  0T=  /3,  then  the  equation  to  8T  will  be 

M- c^)- 

To  find  the  condition  that  (2)  may  touch  (1),  we  proceed  as 
follows : 

Square  each  member  of  (1)  and  multiply  it  crosswise  by  (2), 
we  thus  get 

X     y  _  f/^"\-     (y 

a  "^  ^  "  |W    "^  V^ 

This  may  be  considered  as  a  quadratic  in  \    ]  •,  and  in  order 

that  (2)  may  touch  (1),  it  is  necessary  that  its  roots  be  equal ; 
hence  its  first  member  must  be  a  perfect  square. 
The  equations  to  PT,  Q8  respectively,  are 

-  +  ^  =  1 


a       b 


where  these  meet,  we  have 


the  equation  to  a  line  passing  through  the  origin,  and  through 
the  intersection  of  PP,  Q8^  that  is  to  OV. 


185U.]  GEOMETRY    OF   TWO    DIMENSIONS.  141 

Again,  to  find  the  equation  to  OR^  we  have  since  the  first 
member  of  (3)  has  been  made  a  perfect  square, 

therefore  squaring,  ^  ("  "  -)  =  2^  (jg  -  -^j  ' 
'1      1\         /I      r 


^^Kl--^)+K^-^)='' 


the  equation  to  OR^  which  agrees  with  that  ah-eady  formed  for 
0  V]  hence  0,  i?,  V  are  in  the  same  straight  line. 

6.  A  series  of  circles  pass  through  a  given  point  0,  have 
their  centres  in  a  line  OA^  and  meet  another  line  AB.  From 
ilf,  iV,  the  points  in  which  one  of  the  circles  meets  the  lines 
OA^  AB^  are  drawn  parallels  to  AB^  OA,  intersecting  in  P. 
Shew  that  the  locus  of  P  is  a  hyperbola,  which  becomes  a 
parabola  w^hen  the  two  lines  are  at  right  angles. 

Take  0  as  origin,  OA  as  axis  of  x,  let  the  equation  to  any 

one  of  the  circles  be 

x^  +f  =  2rx (1), 

and  that  to  AB  xcosol  +  ?/siua  =  a (2) ; 

hence  the  coordinates  of  M  are  2/-,  0,  and  the  equation  to  the 

line  through  J/ parallel  to  AB^  is 

xcosa  +  ?/sina  =  2rcosa (3). 

Now  the  circle  in  general  cuts  AB  in  two  points,  either  of 
which  may  be  denoted  by  N.  The  equation  to  the  line  through 
either  of  these  points  parallel  to  OA  (the  axis  of  ic),  will  be  ob- 
tained by  putting  the  ordinate  of  that  point  =  0,  and  therefore 
the  equation  to  the  pair  of  parallels  will  be  obtained  by  elimi- 
nating X  between  (1)  and  (2).     This  gives 

(«  -y  sina)'"*  +  y'  cos' a  =  2r  cos  a  (a  —  3/  sin  a) ; 
.-.  y'^  +  2  (rcosa  — a)  ^sina  +  d^  —  2arcosa  =  0...(4). 

To  find  the  equation  to  the  locus  of  P,  the  intersection  of  (3) 
with  either  of  these  lines,  we  must  eliminate  r  between  (3)  and 


142  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1850. 

(4),  whence  we  get 

y^  +  (iccosa  +  ^sina  — 2a)  ?/sma  +  d^  —  a  (a?cosa  +  y  slna)  =  0, 

or  ^■'(1  +8in'*a)  +  xi/ cosa  sin  a  —  a  [x  cos  a,  +  3^  sin  a)  +  d^  =  0. 

This  is  the  equation  to  the  locus  of  P  which  is  evidently  in 
general  an  hyperbola.  If  however  the  lines  are  at  nght  angles 
a  =  0,  and  the  above  equation  becomes 

y'^  —  ax  +  d^  =  0, 
representing  a  parabola. 

7.  If  from  the  focus  of  a  parabola,  lines  be  drawn  to  meet 
the  tangents  at  a  constant  angle,  the  locus  of  the  points  of  inter- 
section will  be  that  tangent  to  the  parabola  whose  inclination 
to  the  axis  is  equal  to  the  given  angle.     Prove  this  in  any 

manner,  and  shew  that  if  m  be  eliminated  between  y  =  mx  -\ —  , 

and    V  = [x  —  a),    the   result    contains   a  factor   which 

answers   to   the   locus.      Also   explain   briefly   the   origin   and 
signification  of  the  other  factors. 

(a).  Let  4a  be  the  latus-rectum  of  the  parabola,  a  the  iji- 
clination  to  the  axis  of  any  one  of  the  series  of  tangents,  yS  the 
constant  angle  at  which  the  lines  through  the  focus  meet  the 
tangents.  Then  taking  the  focus  as  pole,  and  the  axis  as 
initial  line,  the  equation  to  the  tangent  is 

r  =  acoseca  cosec(^+  a). 
That  to  the  line  through  the  focus  is 

6/  =  7r-(a  +  ;8). 
Eliminating  a,  we  get  as  the  locus  of  the  intersection  of  these 
"^^^  r  =  acosecyS  cosec(^  +  yS), 

representing  the  tangent  whose  inclination  to  the  axis  =  /3. 

/3.    From  the  equation 

m  +  t  . 
y  = [x  —  a) (1), 

we  get  m  [x  —  a  +  ty)  =  —  t{x  —  a)  +  y  ', 


1850.]  GEOMETRY   OF   TWO   DIMENSIONS.  143 

therefore  combiuing  this  with 

y  =  mjc  +  ^^ (2), 

y  —  tix  —  a]  fy  +  X  —  a 

y  =  ^ !^ 1  X  +  — -. -,  a, 

ty  +  X  —  a  y  —  t[x  —  a) 

y— tix—a) ,        ,        ill— tix—a)        ty  +  x  —  a 
-  '^ ^ {x-a)+a]^ ^ ^  +    ^ 


ty  +  x-a^  [ty  +  x  —  a       y  —  t[x—a))^ 

{{x-a)^  +  f}{l+f)      . 
ty  +  x  —  a)  [y  —  t[x  —  a)}  ' 

X  —  a)''  ■+  y'     ty  —  fx  —  a 


^f+[x-ar^^     {[x-aY  +  f]{l+f)      . 
ty  +  x  —  a  i^ty  +  x  —  a)[y  —  t[x  —  a)]^ 


=  0, 


X  —  a  +  ty   '  y  —  t{x  —  a) 
This  is  satisfied  by        y  =  tx  +  -  ^ 

representing  the  locus  found  above. 
It  is  also  satisfied  by 

[x-aY  +  y'  =  % 

which  requires  that  cc  =  «,  ?/  =  0,  representing  the  focus. 

This  would  be  obtained  by  making  m  =  (—1)*  in  equation  (2). 

Its  signification  therefore  is,  that  if  tangents  be  drawn  to  the 
imaginary  branch  of  the  parabola,  got  by  making  x  negative, 
and  lines  be  drawn  through  the  focus  of  the  real  branch  cutting 
these  tangents  at  a  constant  angle,  the  point  of  intersection  of 
these  lines  will  only  be  real  when  the  tangent  to  the  imaginaiy 
branch  of  the  parabola  passes  through  the  focus  of  the  real 
branch. 

8.  Within  the  evolute  of  an  ellipse  is  inscribed  a  similar 
ellipse ;  within  its  evolute  another  similar  ellipse,  and  so  on 
ad  infinitum ;    shew  that  the  sura  of  all  the  areas 


TT 


id'+V') 


4         ah 

Let  mrt,  mh  be  the  semi-axis  of  the  first  inscribed  ellipse, 
then  m  will  be  a  homogeneous  function  of  a  and  h  of  no  dimen- 
sions, and  therefore  the  same  function  of  ma  and  mh ;   hence 


144  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [18oU. 

wt^/,  ni^b  will  be  those  of  the  second,  and  the  sum  of  all  the  areas 

TTuh 


1  -  m' ' 

we  have  therefore  only  to  find  the  value  of  m. 
Now  the  equation  to  the  evolute  of  the  ellipse  is 

{axf+{hyf  =  {d^-hi (1). 

In  order  that  this  may  touch  the  ellipse 

\ina)        \mh)  

they  must  have  a  common  tangent  at  a  common  point. 
Now  the  equation  to  the  tangent  to  (1)  at  [xy]  is 

In  order  that  this  may  touch  (2)  at  {xy)^  we  must  have 
'a^i         1  X 


(S^^.-(3..  =  K-0-'. 


2  7 


27,a   J 


X  ina 

'   •  2    2  2  z3j  7 

ma       a  —  0 
y^    _      mW' 

therefore,  by  (2),     m  =  ^r-^  5 

therefore  if  S  equals  the  sum  of  all  the  areas, 

Trab 


S  = 


a'-  b'\' 


a'  +  ¥ 
4         ah 


1850.]  GEOMETRY   OF   TWO    DIMENSIONS.  145 

9.  Find  the  points  ^,,  A^^  A^...A^^^_^,  A^^^  in  a  parabola,  such 
that  the  tangents  at  these  points  are  parallel  to  the  focal  dis- 
tances SA^^^^  SA^,  SA^y..SA^_^^  ^^„t-\-)  respectively. 

TiQl  A^A^KV  [^g.  71)  represent  the  parabola,  A''/S'X  its  axis, 
A^T^^ ...  A^T^^  the  tangents  at  ^„...^,,,  respectively,  and  let 
lA^SX  =  a^,  then  lA^T^X  =  ^a^.  Hence,  by  the  conditions  of 
the  problem,  we  must  have 

W=^m  (1), 

K  =  «.    (2), 

K  =  a,-, (''), 


^m-l   =  am-2 (w-1). 


The  last  equation  will  be 

TT  +  ia,,,  =  «„,_, (m), 

since  this  will  satisfy  the  condition  of  A^^^T^^  being  parallel  to 
SA^^^_^^  and  it  is  manifestly  inconsistent  with  the  preceding 
equations  to  have  ^a^^^  =  a^^^_^. 

Hence,  multiplying  generally  equation  (?')  by  2'""^,  and  add- 
ing all  equations  thus  formed,  the  quantities  a^,  a^,  ...  «,,_  ,  dis- 
appear, and  we  get 

^  +  i««,  =  2"'-'a,„ ; 
-      27r 

•■•      «m    -    2'"    _    1  • 

whence     .,  =  ,-J^  , 

2'-+«_. 
and  generally,     a,.  =  ^„.  _  ^  , 

and  the  positions  of  the  points  A^^  A^...A^  are  determined. 

10.  From  the  focus  of  an  ellipse  lines  are  di-awn  to  any  four 
points  in  the  curve,  and  the  reciprocal  of  each  line  Is  multiplied 
by  the  sines  of  half  the  angles  between  any  two  of  the  remaining 
lines ;  prove  that  the  sum  of  the  first  and  third  of  these  products 
taken  in  order  is  equal  to  the  sum  of  the  second  and  fourth. 

L 


146  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

Let  L  =  the  latus-rcctum,  e  the  eccentricity  of  the  ellipse ; 
then  Its  equation,  refeiTcd  to  the  focus  as  origin  and  the  axis- 
major  as  prime  radius,  will  be 

1  _  1  —  e  cos^ 

and  let  ^,,  6^^  6^^  6^  be  the  angles  between  the  axis  and  the 
successive  lines,  r^,  ;-^,  rg,  r^  the  lengths  of  the  lines. 

Then  —  sm  -2-— — -*  sm  ^— — ?  sm  -^ ^ 

y-j  2  2  2 

2  ^0-0,     0-6.     (^-f> 

=  ^  (1  —  e  cos  ^j)  sin  -^-— — *  sm  -^-— — ?  sm 


Z  ^  >^  2  2  2 

=  ^  (1  -  e  cos  ^J  (sin  [6^  -  e,)  +  sin  (^,  -  ^J  +  sin  (^3  -  ^,)}. 

O-        .,         ,  1         .         ^,     -    ^.       •         ^.>    -    ^4       •  ^4-^1 

bimilarly     —  sm  ^-- —  sm  -^— - —  sm  -^-— — - 
•'      r^  2  2  2 

=  ^  (1  -  e  cos^3)  {sin  (^,  -  e^)  +  sin  (^,  -  ^J  +  sin(^^  -  ^J]; 

therefore  if  ^8'  denote  the  sum  of  these  two  quantities,  and  If  the 
symmetrical  fmiction  sin  {6^  —  6^  +  sin  {6^  —  6^  +  sin  (^3  —  6^ 
+  sin  (^.^  -  6^)  -}-  sin  [d^  -  ^J  +  sin  [6^  -  6^  be  denoted  by  <^, 
and  if  we  put  c^  for  cos^^,  s^  for  sin  ^j,  &c., 

+    ^3  k^l  -  ^1^.2  +  ^4^2  -  ^^4  +  ^'^1^4  -  ^4^1)}] 
=    ^  [0  -  &A[^-h)+  ^-f^i^-^l)  +  C3^4(«l-  «2)+^4^lK-*3)}]- 

Similarly,  if  S'  be  the  sum  of  the  second  and  fourth  products, 
^S"  will  be  obtained  from  S  by  writing  6,^  for  6^^  0^  for  ^.^,  6^  for  ^3, 
^j  for  6^ ;  hence  (/),  which  Is  a  spmuetrical  function,  will  remain 
unchanged,  and 

whence  It  appears  that  S  =  /S", 

or  the  sum  of  the  first  and  third  of  these  products  is  equal  to  the 

sum  of  the  second  and  fourth. 


1851.]  GEOMETRY    OF    TWO    1)1  .M  KXSloNS.  147 

11.  If  lines  be  drawn  tlirough  any  two  of  the  points 
A^  i?,  Cy..  and  other  lines  through  any  two  of  the  points 
f/,  i,  Cy..  all  in  one  plane,  prove  that  the  intersections  of  AB 
with  «7>,  of  AG  with  «c...,  will  all  lie  in  one  straight  line,  pro- 
vided that  the  lines  through  the  intersections  of  any  two  of 
the  first  series  of  Hues  and  the  corresponding  intersections  of 
the  second  series  all  pass  through  the  same  point. 

Conceive  the  points  -4,  i?,  C,  ...  to  lie  in  one  plane,  and 
((,b,Cy..  in  another;  then,  since  the  lines  joining  the  intersec- 
tions of  any  two  of  the  first  series  of  lines  and  the  corresponding- 
intersections  of  the  second  scries  all  pass  through  one  point, 
Aa^  Bbj  Cc...  all  pass  through  one  point  0. 

Now  consider  any  quadrilateral,  as  ABah^  whose  angular 
points  are  any  two  points  of  the  first  series,  and  the  correspond- 
ing two  of  the  second  series.  Since  the  lines  Aa^  Bb  intersect, 
they  are  in  the  same  plane,  therefore  also  AB^  ab^  are  in  the 
same  plane,  and  must  therefore  intersect,*  and  their  intersection 
must  manifestly  lie  in  the  line  of  intersection  of  the  planes 
ABC...,  abc...  Similarly  the  intersection  of  any  other  pair  of 
lines,  as  A  C,  ac,  lies  in  that  line. 

Hence,  if  we  suppose  the  planes  ABC...,  abc...,  to  be  indefinitely 
nearly  coincident,  the  proposition  enunciated  follows  at  once. 

1851. 

1.  Having  given  a  focus  and  two  tangents  of  a  conic  section, 
shew  by  means  of  reciprocal  polars,  or  otherwise,  that  the 
chord  of  contact  always  passes  through  a  fixed  point. 

Let  a  circle  be  described  passing  through  two  fixed  points, 
A,  B,  and  let  F  be  the  intersection  of  the  tangents  at  A,  B. 
The  locus  of  P  will  be  a  fixed  straight  line,  pci-pendicular  to  and 
bisecting  AB. 

Now  take  the  polar  reciprocal  of  this  system  with  respect 
to  any  fixed  point  S.  The  reciprocal  of  the  circle  will  be  a 
conic  section  whose  focus  is  S,  and  which  has  two  fixed  tangents 
(the  reciprocals  of  A,  B).     Hence  the  reciprocal  of  P,  which  is 

*  If  these  lines  happen  to  be  parallel  we  may  still  consider  them  as  inter- 
secting in  a  point  infinitely  distant. 

L2 


148  SOUTTIOXS   OK   SENATE-HOUSE    PROBLEMS.  [1851. 

tli(^  chord  of  contact  of  these  tangents,  will  always  pass  through 
a  fixed  point,  the  reciprocal  of  the  locus  of  P. 

2.  Shew  that  there  will  be  two  pairs  of  equilateral  hyper- 
bolae which  pass  through  two  given  points  -4,  B^  and  touch  two 
given  straight  lines,  and  that  the  chords  of  contact  of  each  pair 
meet  in  AB^  and  are  equally  inclined  to  AB.^ 

Take  the  middle  point  of  AB  as  origin,  AB  as  axis  of  a*, 
let  h^  —  h^  be  the  abscissse  of  A^  B,  respectively,  and  let  the 
equations  to  the  two  given  tangents  be 

5  +  f_l  =  0,     ^+1-1=0, 

and  that  to  their  chord  of  contact  -  +  ^—1=0, 

a      p  ' 

where  a,  /9,  are  indeterminate  parameters. 

Then  the  equation  to  a  conic  section  touching  the  two  given 

lines  may  be  written  mider  the  form 

M->)(M-')=^(M-' 

\  being  an  indeterminate  parameter. 

Two  equations  for  the  determination  of  the  three  arbitrary 
quantities  X,  a,  yS,  are  given  by  the  conditions  of  its  passing 
through  u4,  B.     We  thus  get 

^o(^')=Ks-y «- 

a        J   \a        J  \a. 

The  third  equation  is  given  by  the  condition  of  the  curve 
being  an  equilateral  hyperbola.  In  order  that  this  may  be  the 
case,  it  is  necessary  that  the  sum  of  the  coefficients  of  x^  and 
y-*  =  0.     This  gives 

—  -  ^.  +  777  -  3i  =  0 3  . 

aa       a.        bo       p 

*  A  shorter  solution  of  this  problem,  due  to  Mr.  Gaskin,  will  be  found  in 
the  Appendix. 


1851.]  GEUMETUY    OF    TWO    DlMENS-lONS.  149 

Combining  (1)  and  (2),  we  get 

^h  —  aV       (h  —  a)  (h  —  a) 


Kh  +  aj        (A -f  a)  {h  +  a) ' 

This  is  a  quadratic  for  the  determination  of  a.     Let  a,,  a„  be 
its  roots. 

Subtracting  (1)  from  (2)  and  eliminating  \  by  (3),  we  get 


1      1\ 

n      i\ 

2 

/I         1  ■ 

,?  +  w) 

\a      a'J 

a 

\aa'       bh\ 

Hence,  for  each  vahie  of  a  there  will  be  two  of  yS,  equal  and 
of  opposite  signs.  Let  them  be  denoted  by  ^8^,  —  ^,,  ^^^  —  /S^ 
respectively,  then  we  get  two  pairs  of  equilateral  hyperbolae, 
whose  chords  of  contact  respectively  are 

-  +  !-  =  1,  --i  =  1 5. 

It  is  easy  to  see  that  the  lines  represented  by  (4)  intersect 
in  AB  (the  axis  of  x)  and  are  equally  inclined  to  AB.  The 
same  will  be  the  case  with  the  pair  of  lines  denoted  by  (5). 
Hence  the  chords  of  contact  of  each  pair  meet  in  AB^  and 
are  equally  inclined  to  AB. 

3.  If  from  a  point  of  an  ellipse  a  line  be  drawn  to  the  ex- 
tremity of  each  axis,  and  a  parallel  to  the  same  axis  be  drawn 
through  the  point  in  which  such  Hue  meets  the  other  axis,  the 
locus  of  the  intersection  of  these  parallels  is  an  equilateral  hy- 
perbola. 

Trace  the  coiTesponding  positions  of  the  point  on  the  hy- 
perbola, and  the  point  on  the  elhpse. 

Let  a,  J,  be  the  semi-axes  of  the  ellipse,  take  the  axes  of 
the  cui've  as  coordinate  axes,  and  let  a  cos  a,  h  sin  a,  (a  being  a 
variable  parameter)  be  the  coordinates  of  the  point  tlu'ough 
which  the  lines  are  drawn.  The  equation  to  the  line  drawn 
through  this  point  to  the  extremity  of  the  axis-major,  is 
X  —  a  cosa  y  —  ^  sin  a 
a  cosa  —a  h  sina      ' 


150  S<lMTIttN8    OF    SKNATK-IIUUSE    PROBLEMS.  [1851. 

X  —  a  cosa      y  —  h  sin  a 

or    ^-j +''  =0; 

a  sill  ^  a  h  cos^a 

X       a       11    .    OL  a 

.'.     -  cos-  +  Y  sill-  =  cos  -  . 

a        2       h        2  2 

Where  this  meets  the  axis-minor, 

ij  =  hcot^a. (1), 

which  is  therefore  the  equation  to  the  line  through  this  point, 
parallel  to  the  axis-major. 

Again,  the  equation  to  the  line  drawn  through  («  cosa,  b  sin  a) 
to  the  extremity  of  the  axis-minor,  is 

y  —  h  siiia  _x  —  a  cosa 
Z»sina  —  h  a  cosa      ' 

y  —  h  cos  (I  TT  —  a)       x  —  a  sin  ( 2  "^  —  a)  _ 
h  cos(-2  7r  —  a)  —  h  asin(^7r  —  a) 

whence,  by  an  investigation  similar  to  the  above,  it  is  seen  that 
the  equation  to  the  line  drawn  parallel  to  the  axis-minor  through 
the  point  where  this  meets  the  axis-major,  is 

X  =  acot(j7r  —  |-a) 

=«3^i (^i- 

Eliminating  a  between  (1)  and  (2),  we  get 

y  +  b 

y-h' 
or    xy  =  ay  +  hx  +  ab^ 

or    {x  —  a)  [y  —  b)  =  2aJ, 

as  the  equation  to  the  locus,  which  is  evidently  an  equilateral 
hyperbola,  whose  asymptotes  are  the  tangents  to  the  ellipse  at 
A  and  B. 

Let  ABA'B  (fig.  72)  be  the  ellipse,  P  the  point  (a  cosa, 
b  sin  a) ;  then  from  the  figure  it  appears  that  p  is  the  point 
on  the  hyperbola  corresponding  to  P:  one  branch  of  the  hyper- 
bola is  described,  while  P  moves  from  A  io  B'^  the  other  branch 
while  2>  moves  round  through  BA'B'A. 


1851.]  GEOMETRY   OF   TWO    DIMENSIONS.  151 

4.    Dctenniue  the  values  of  a',  m\  n\  such  that  the  relation 
{{x-af  +  /}*  =  m  (x^  +  y'y-  +  n 
may  be  equivalent  to  the  relation 

[{x  -  a)"  +  y'Y-  =  m  {x'  +  y'^Y  +  n. 
The  transfonnation   fails  (1)   in  the  case  where  the  curve 
represented  by  the  given  equation  is  a  conic  section,  (2)  has  a 
double  point. 

(a).    The  equation 

{{x-a'Y  +  f]^  =  in  {x'  +  f)^  +  n\ 

when  transformed  to  polar  coordinates,  and  rationalized,  becomes 

r^  —  2a.'r  cos^  +  a.'^  =  (?«';•  +  ?«y, 

or    (1  -m")r'  -  2a'r  cos^  -  2m'n'r  +  a'  -  ?i"  =  0. 

The  equation 

{(a; -a)'  +  /}*  =  m  {x'+iff-  +  n, 
similarly  transformed,  becomes 

(1  -  7n')  r^  -  2a.r  cos 6  —  2mnr  +  ai'  —  n"  =  0. 
In  order  that  these  equations  may  be  identical,  the  coefficients 
of  r^,  r  cos  ^,  r,  must  bear  the  same  ratio  to  one  another  as  the 
constant  temis, 

1  —  ni''      OL       m'n       a^  —  n^ 


1  -  m^ 

a 

mil        a'"*  - 

-n' 

lation 

a 

^ 

a"  -  n"' 

Q                    ■> 

a 



a^  -  71^ 
1-^ 

we  get 

a' 
a 

= 

-? 

,    »" 

'-^' 

1  - 

1  -  in' 

I  - 

n' 

^ 

1  -  m'  ' 

^' 

152  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 


Also  wc  have 


n 

a 

mil            7)1 

a 

'e 

— 

^ 

,     or  — 

—  ^   i 

a 

vin             VI 
n 

71    ' 

a 

• 

m' 

^ 



a  ' 

and 

n' 

= 

m, 

a 

a 

iiice 

a 

1  -  nf  ' 
1  a'  -  n' 

• 

a! 

^^ 



a  1  —  m^ 

711  a"'*  —  n^ 

n 

^ 

a  1  —  nf  '" 

(1), 


(2), 

(3). 

The  required  values  of  a',  m',  w',  are  determined  from  (1),  (2), 
and  (3). 

(y8).  The  given  equation,  when  rationalized,  will  in  general 
be  of  the  fourth  degree  in  x  and  y.  In  order  therefore  that 
it  may  represent  a  conic  section,  it  is  necessary  that  the  terms 
of  a  degree  higher  than  the  second  should  disappear  of  them- 
selves from  the  rationalized  equation,  which  requires  that 
7)1  =  ±  1.     If  this   condition   be   satisfied,   the   given   equation 

DGCOmGS 

{{x-ay  +  f]i±  {x'+7/'y  =  7i, 

shewing  that  one  of  the  foci  of  the  conic  section  is  the  origin, 
that  the  coordinates  of  tlie  other  are  a,  0,  and  that  the  axis- 
major  =  71. 

The  transformed  equation  must  therefore,  since  it  represents 
a  conic  section,  take  the  form 

{{x-a'Y  +  7/}i±{x'+f)i  =  7l\ 

which  gives  a',  0  as  the  coordinates  of  the  second  focus,  7i  as  the 
axis-major.     Hence  we  must  have 

a'  =  a,     w'  =  ;«, 
and  the  transformation  fails. 


1851.]  GEOMETRY   OF   TWO   DIMENSIONS.  153 

The  transfonnation  also  fails  if  n  =  a,  for  we  then  get  m'  =  1 , 
a  =  0,  n  =  0,  and  the  transformed  equation  becomes  an  identity. 
In  this  case,  the  given  equation,  transformed  into  polar  coor- 
dinates, becomes 

(1  -  vf)  r  -  2  (a  cos  ^  +  ma)  =  0  ; 

whence  we  see  that  if 

cos^  =  —  m^ 

we  have  r  =  0 ;  but  the  equation  cos  0  =  —  m  is  satisfied  in 
general  by  two  values  of  ^,  whose  sum  =  27r ;  hence  the  origin 
is  a  double  point. 

If  therefore  the  curve  have  a  double  point,  the  transfonnation 
fails  agam. 

5.  If  0  be  the  centre  of  a  reflecting  circle,  Q  a  radiant  point, 
and  the  line  from  Q  to  0  produced  to  meet  the  circle  be  con- 
sidered as  the  axis,  then,  if  a  be  the  radius,  u  the  distance  QO^ 
0  the  inclination  to  tlie  axis  of  the  radius  through  the  point  of 
incidence  of  any  ray,  and  <f>  the  inclination  to  the  axis  of  the 
reflected  ray, 

pcos(f)  =  acosd  +  wcos2^,     psmcj)  =  asm0  +  usm20j 
where  p  =  {a'  +  u^  +  2au  cos  0Y  is  the  length  of  the  incident  ray. 

Let  P  (fig.  73)  be  the  point  of  incidence  of  any  ray,  M  the 
point  in  which  the  reflected  ray  cuts  the  axis.     Let 

and  draw  PN  perpendicular  to  the  axis.     Then 

QN=  /3  cos  (<^  -  21/r)  =  QO  +  ON  =  u  +  «cos^, 
PN  =  p  sin  (0  —  2\|r)  =  a  sin  0. 

Again,    yjr  =  (ji  —  0, 

.'.  p coa{'20  —  (j>)  =  u  +  acos0  (1), 

p8in(20-<^)  =  asin^ (2): 

(1)  cos 2^  +  (2)  sin  2^  gives 

p  cos<^  =  a  cos  0  +  u  cos  2^, 


154  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1851. 

(1)  sill 2^  —  (2)  cos2^  gives 

psincf)  =  aamO  +  usm20j 
the  required  equations. 

6.  Using  the  notation  of  the  last  question,  and  assuming  the 
tiTith  of  the  theorem  stated  therein,  shew  that  if  fi-om  the  point 
of  incidence  of  each  ray  there  be  drawm,  in  a  direction  opposite 
to  that  of  the  reflected  ray,  a  line  equal  in  length  to  the  incident 
ray,  the  locus  of  the  extremities  of  these  lines  is  a  curve  cutting 
the  lines  at  right  angles,  and  the  equation  of  which,  referred  to 
the  radiant  point  as  origin  and  the  axis  QO  SiS  axis  of  x^  is 

x^  +  y^  —  2ux  =  2a{x^ -i- y^)^. 

Shew  that  the  origin  is  a  double  point,  and  trace  the  curve : 
shew  also  that  the  equation  may  be  expressed  in  the  form 

{{x  —  af  +  /]*  =  m  [x'-\-y^)^  +  n. 

'  (a).  Produce  MP  to  i?,  making  PR  =  QP^  then  we  have 
to  find  the  locus  of  R.  Let  x^  y  be  its  coordinates,  then  we 
readily  see  that 

x=  QN  -\-  pcos<f) 
=  u  +  acosd  +  acosd  +  ncos20 
=  u{l  +  cos2d)  +  2a  cos  ^, 
and  y  =  PN  +  psin^ 

=  asin6'  +  asin^  +  ?/ sin2^ 
=  M  sin  2^  +  2a  sin  6 : 

hence  -  =  tan^,    and  RQ  is  parallel  to  PO, 


and  [x- 

-uY 

+  f 

=  u^  +  4«''  +  4:au  cos  6, 

.-.  x'  + 

■f- 

-2ux 

X 
■—  A  /y      1     A  /ytt 

x'  +  f  - 

2ux 

+  li' 

x'              .    ■>         .                 X 

^  + 

w.^ 

x'' 

2     ,         -2    —    *<^*      "T    ^:''"     /     2     ,         2^ 

X  +y                 G-^  +,y . 

0.'^+/' 

1851.]  GEOMETRY    OF    TWO    DIMENSIONS.  155 

.'.  x^  +  }f  -2ux  =  2rt(x'+y')*, 
the  required  equation  to  the  curve. 

(y8).  This  equation,  transformed  to  polar  coordinates,  becomes 
r  —  2u  0,09,6  =  2a, 

.-.   r  =  211CO&0  +  2a (1). 

Hence  the  radius  vector  of  the  curve  exceeds  by  2a  the 
radius  vector  of  a  circle  which  passes  through  the  pole,  one 
of  whose  diameters  is  prime  radius,  and  whose  radius  =  u.  If 
therefore  we  draw  such  a  circle,  and  produce  0-4*,  the  radius 
vector  of  any  point  A  in  it  to  i?,  making  AB  =  2a,  the  locus 
of  R  will  be  the  required  cm've. 

The  cm-ve  will  pass  through  the  origin  when  cos^  = , 

which  condition,  if  a  <  u,  is  satisfied  by  two  values  of  0,  one  less, 
the  other  greater  than  tt.  Hence  if  a  <  w,  i.e.  if  Q  be  outside 
the  circle,  two  branches  of  the  curve  pass  through  the  origin, 
which  is  therefore  a  double  point. 

When  ^  =  0,  /■  =  2  {u  +  «),  and  when  6  =  tt,  r  =  —  2 [u  —  a) ; 
hence  the  curve  will  have  the  form  represented  in  fig.  (74),  where 

Qq  =  2{zi  +  a),      Qq  =2{u-a).-f 
Again,  l  QRM=  L  0PM  (since  QR  is  parallel  to  OP)  yfr  =  (f>-0, 
sm<f>  cos^  —  cos^  sin^ 


tSinQRM  = 


COS0  cos^  +  sin<^  sin^ 
u  sin  0 


a  +  Mcosi 
by  the  result  of  question  5. 


*  Tlie  line  OA  must  ahvaj's  be  produced  in  the  positive  direction  of  the 
radius  vector,  therefore  when  0  >  .Vtt,   OA  must  be  produced  backwards. 

t  This  is  the  form  of  the  figure  when  Q  is  outside  the  circle  :  if  it  be  witliin 
it,  the  curve  docs  not  pass  through  the  origin,  and  the  loop  Qq'  docs  not 
appear.     The  origin  will  then  be  a  conjugate  point. 


15G  SOLUTIONS  OF   SENATE-HOUSE   PROBLEMS.  [1851. 

And  if  L  be  the  angle  between  the  radius  vector  and  tangent, 

1  flu  u  sin  0 

cott  =  — T7i  =  — 


L  dO  a  +  u  cos  6  ' 

=  tan()i?J/. 
Hence  the  curve  cuts  the  lines  PR  at  right  angles. 

(7).    The  equation 

[{x  -  a)'  +  y'Y-  =  m  {x'  +  y^  +  n 
is  equivalent  to 

(1  —  m^)r^  —  2mnr  —  2ar  cos^  +  a^  —  w^  =  0, 
and  this  coincides  with  (1),  if 

-2  =  «?     -. -2  =  ^1     a.  -  n  =0] 


1  _  ,n^  -  "'     1  _  ni' 


a  u  —  a 

m  =  -  .     71  =  a  = 


,/<-  —  >*—  , 


u  u 

therefore  (1)  is  equivalent  to 

,2  2^  2  N  J 


w  —  a\        ,J^      u  ,  .,       .,,,       It  —  a 


'■^\* 


^ — :— +3/    =-i^"+rf  + 


) 


M     /         j        a  u 

which  is  in  the  required  fonm. 

7.  Given  the  centres  of  three  circles,  each  of  them  touching 
the  other  two  externally,  determine  the  radii. 

How  many  systems  of  circles  are  there  when  the  centres 
are  given,  but  the  circles  touch  externally  or  internally  at 
pleasm'c  ? 

Let  a,  &,  c,  be  the  distances  between  the  given  centres ;  then 

'•2  +  ''3  =  ^) 

^3  +  '\  =  h 

r^  +  r.^  =  c, 

b  +  c  —  a 


similarly  r^  = 


2  ' 

c  +  a  —  b 

2  ' 

a  +  b  —  c 


which  determine  the  radii. 


1851.]  GEOMETRY    OF   TWO    DIMENSIONS.  157 

If  the  circles  touch  internally  or  externally  at  pleasure,  there 
will  be  four  systems.  For  the  circle  described  with  any  one 
point  as  centre  may  include  the  other  two,  which  must  touch 
each  other  externally,  thus  giving  three  systems.  Or  they  may 
all  touch  externally,  giving  four  systems  in  all. 

8.  The  locus  of  the  point  from  which  two  given  circles 
subtend  equal  angles  is  a  circle. 

Let  -4,  A'  (fig.  75)  be  the  centres  of  the  two  given  circles, 
P  a  point  from  which  the  circles  subtend  equal  angles.  Draw 
the  tangents  P T,  Ft  to  the  circle  whose  centre  is  A ;  PT\  Ft'  to 
that  whose  centre  is  A'.  Join  P/1,  PA\  AT^  At,  A'T\  A't'. 
Then  LTFt  =  AT  Ft'. 

And  AF,  A'F  respectively  bisect  the  angles  PPf,  TFt', 

.-.  z  AFT  =  L  AFT: 

also  the  right  angle  ATF  =  \k\e,  right  angle  A'T'F',  therefore 
the  triangles  TAF,   TA'F  are  similar,  therefore 

AF'.A'F::  AT:  AT (1). 

Divide  AA  in  0,  so  that  AO  \  A 0  ::  AT :  AT  take  0  as 
origin,  A OA  as  axis  of  x.  Let  AO  =  a,  A 0  =  a',  and  let 
X,  y  be  the  coordinates  of  P.     Then  by  (1) 

...  d'%x^a)^^f]=d^[{x-d)'^f], 
or  (a  —  a')[x^  -\-y'^)  —  'iaax  =  0  ; 
shewing  that  the  locus  of  P  is  a  circle  passing  through  0. 

9.  The  lines  joining  the  corresponding  points  of  two  similar 
and  similarly  situated  figures  in  the  same  plane  intersect  in  a 
point. 

All  sections  of  a  conical  surface  of  any  degree  by  parallel 
planes  are  similar  and  similarly  situated  figures,  and  every  gene- 
rating line  passes  through  coiTCsponding  points.  Hence,  con- 
versely, the  lines  joining  coiTCsponding  points  of  two  similar  and 
similarly  situated  figures  in  parallel  planes,  pass  through  one 
point    (the   vertex   of  the   conical   sm'face   of  which   they  are 


158  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1851. 

sections).      Let    the    planes    l)e    now    made    indefinitely    nearly 
coincident,  and  the  proposition  enunciated  follows  at  once. 

10.  Given  any  three  of  the  four  lines  Ox^  Oy^  0])^  Oq^ 
(fij^.  76),  the  foiu'th  may  be  determined,  such  that  if  «,  h  be  the 
points  in  which  a  line  tln-ouf2,h  a  point  Q  in  Oq  intersect  Ox^  Oy^ 
and  n\  h'  the  points  in  which  another  line  through  the  same 
point  Q  intersects  Ox^  Oy^  the  point  of  intersection  of  the  lines 
ah'  and  ah  lies  on  the  line  Oj). 

Let  II  =  0,  V  =  0,  be  the  equations  to  any  two  lines  passing 
through  the  point  0,  and  let  u  —  \v^  u  =  Xv,  u  =  X^r,  w  =  \v, 
be  the  equations  to   Ox,   Oy,   Oq^j   Oq,  respectively.     Also  let 
It)  =  0  be  the  equation  to  Qa,  and  n  —  \ii  —  fiic  =  0  that  to  Qa. 
Then  the  equation 

a{u  —  \v)  —  [u  —  \gV  —  [xw]  =  0, 
where  a  is  a  disposable  parameter,  represents  a  line  passing 
through  a.     In  order  that  this  may  pass  through  h,  the  above 
equation  must  be  identical  with 

l3{u-\v)-w  =  0, 
y8  being  also  a  disposable  quantity.     In  order  that  these  equa- 
tions may  be  identical,  we  must  have 
a  —  1       a\,.  —  \a 


and  (Xy  —  \j){u  —  \v)  —  {\y  —  \.) («  —  \v  —  fiw)  =  0, 
or  {\ - \){u - Xv)  +  (\ -X^)fiw  =  0, 
is  the  equation  to  ah.     Similarly 

(\  -  \)  (^  -  K^)  +  {\  -  \)  /^^^'  =  0, 

is  that  to  ah'.     Where  these  intersect,  we  have 

(X^  -  \,)  [u  -  Xv)  +  (X  -  \)  (u  -  Xv)  =  0, 

or   {X^  +  Xy-2\)u  +  X,{X^  +  Xy)v  =  0 (1). 

In  order  that  this  may  lie  in  the  line  0/j,  whose  equation  is 


ti  -  X^v  =  0 (2) 


1851.]  GEOMETRY   OF   TWO    DIMENSIONS.  159 

(1)  and  (2)  must  be  identical ;  hence 

\  i\  +  \-  2\)  +  \  {\  +  \)  =  0, 

or  {\  +  X,]{\  +  \)  -  2\X,  =  0, 
an   equation   from   which,   when  any   three   of  the   quantities 
X^,  X^,   \  ,  X^,    are   given,  the  fourth  may  be  determined,   so 
that  when  any  three  of  the  lines  Ojc,  Oy^  Op^  Oq  are  given,  the 
fourth  may  be  determined  so  as  to  satisfy  the  required  conditions. 

11.  The  radii  vectores  from  the  focus  of  a  conic  section  to 
two  points  of  the  curve  make  equal  angles  with  the  line  drawn 
from  the  focus  to  the  point  of  intersection  of  the  tangents  at  the 
two  points. 

Let  a,  (B  be  the  angles  which  the  radii  vectores  respectively 
make  with  the  axis-major,  then  the  polar  equations  to  the 
tangents,  referred  to  the  focus  as  pole  and  the  axis-major  as 
prime  radius,  will  be 

1      2 

-  =  -  {^cos^  +  cos(^  — a)}, 

1       2 

-  =  y  {e  cos^  +  cos  [6  - y8)|. 
r       i 

Where  these  meet,  we  must  have 

which  will  be  the  equation  to  the  line  through  the  focus  and  the 
intersection  of  the  tangents,  which  evidently  bisects  the  angle 
between  the  radii  vectores.     Hence  the  proposition  is  tnie. 

For  a  demonstration  of  this  theorem  by  the  method  of  re- 
ciprocal polars,  see  Salmon's  Conic  Sections^  chap.  xiv. 

12.  If  two  triangles  be  circumscribed  about  a  conic  section, 
their  angular  points  lie  in  another  conic  section. 

Let  u  =  0,  V  =  0,  t«  =  0,  be  the  equations  to  the  sides  of  one 
triangle,  and  let  the  sides  of  the  other  triangle  respectively 
opposite  to  these  be  represented  by 

?/  +  &,t'  +  Cj?c=0.,.(l),    a^«  +  t'  +  r,?r=0...(2),    a^u  +  \v+ir=0...{?,). 


160  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Then,  since  these  two  triangles  are  circumscribed  about  a 
conic  section,  it  will  follow  that  if  (ly^'J  denote  the  line  joining 
the  intersection  of  v  =  0  and  (3)  with  that  of  i«  =  0  and  (2), 
with  similar  notation  for  the  other  corresponding  lines,  {v^io^), 
{^o^v^),  (?//',)  all  pass  through  one  point. 

Now  the  equation  to  (^'g^t'J  is 


V  10 

w  +  -  +  —  =  0 
that  to  2v,u„  IS   —  -\-  V  +  —  =  0 

U  V 

to  t/.-y,,  — I h  w  =  0 


(A). 


The  elimination  of  w,  r,  lo  between  these  equations  would 
give  the  necessary  condition  that  the  three  lines  denoted  by 
them  should  pass  through  one  point,  or  that  the  two  triangles 
should  be  cii'cmnscribed  about  a  conic  section. 

Now  the  equation  to  any  conic  circumscribing  the  triangle 
(123)  can  be  put  into  the  form 

a  [a^u  -\-v-\-  c^w)  {a^u  +  h^v  +  w)  +  ^  [a^u  -\-  b^v  +  w)  [u  -{■  h^v  +  c^w) 

+  'y[u-\-  l\v  +  c^w]{a^u  -{- V -\- c,^ic)  =  0 (4). 

Here  the  coefficient  of  li^  is  proportional  to 

«  +  -+-, 
ofv^to  ^  +  /S  +  f , 

n      -2.  .        ^  /3 

of  10    to h  —  +  7. 

c       c 

^1  2 

Hence,  if  we  give  to  a,  yS,  7  respectively  the  values  which 
w,  V,  w  have  at  the  intersection  of  the  lines  (A),  each  of  these 
coefficients  will  vanish,  and  equation  (4)  will  be  reduced  to  one 
involving  vw^  wu,  uv  only ;  It  will  therefore  also  represent 
a  conic  circmnscribing  the  triangle  whose  sides  are  u  =  0, 
V  =  Oj  10  =  Oj  and  consequently,  if  two  triangles  be  circum- 


1851.]  GEOMETRY    OF   TWO    DIMENSIONS.  161 

scribed  about  a  conic  section,  their  angular  points  lie  in  another 
conic  section.* 

From  the   above  proof  it  is  not   difficult   to  see   that  the 
converse  (which  is  also  the  reciprocal  theorem)  is  true. 

13.    If  the  angles  ^,  cj)'  are  connected  by  the  equation 
cos)u,  =  cos^  cos<^'  —  sin^  sin^'(l  —c^  sin"''/i)-, 
and  sin0,  sin^'  arc  the  abscissae  of  points  on  an  ellipse,   the 
semiaxes  of  which  arc  1,   (I— c'^)^,  then  the  tangents  at  these 
points  meet  in  a  point,  the  locus  of  which  is  an  ellipse  confocal 
with  the  given  ellipse. 

Let  f ,  r)  be  the  coordinates  of  the  intersection  of  the  tangents, 
then  the  equation  to  its  polar  is 

1  —  c* 
Let  sin^,  (1  —  o'')*  cos^,    be  the   coordinates   of  the   points 
where  this  line  meets  the  given  ellipse,  then 
^  .    _       77  cos  ^ 

the  roots  of  this  equation  in  6  are  <^,  ^'. 
It  may  be  written  in  the  forai 

(|sin^-ir  =  ^^(l-sin^^); 
1  -' 


.•.    sm<p.sm(p  =  .] 

^  ^  1  -  c" 
Again,  it  may  be  written  in  the  form 

(  t;  cos  ^  )         ^, . ,  ,  ^. 

1-r  . 


cos  ^. cos  <^'  = 


r  +  ^ 


•  Another  solution  of  this  and  of  several  cognate  problems,  will  be  found 
in  a  paper  by  Mr.  Heam,  in  the  Cambridge  and  Dublin  Mathematical  Journal, 
vol.  IV.  p.  2G5,  entitled  "  Singular  Application  of  Geometry  of  Three  Dimen- 
sions to  a  Plane  Problem." 

M 


1G2  SOLUTIONS   OF    SKXATE-HOUSE    PKOBLEMS.  [1851. 

therefore,  Ity  the  given  equation, 

cos/,  (f  4-  ^)  =  1  -  f  -  (]  -  ^^,)  (1  -  c^  sin»S 

or  ^'  (1  -t-  cos /a)  +  —^  {cos  //  -  (1  -  c'  sin»^j  =  1  -  (1  -  c'  sin">)S 

the  equation  to  the  locus  of  (^,  t;,)  which  is  therefore  an  ellipse, 
the  squares  of  wliosc  semiaxes  are  respectively 

l-(l-c''sin»^  l-(l-c''sinV)^     , 

1  +  cos  /x         '   ^  '   cos  /i  —  (1  —  c"*  sin'V)- ' 

1  —  cos/A  —  (1  —  cos/i)  (1  —  c*^  sin'^yu-)*  ^ 
or  ;  g  , 

sm  /i 

1  —  c''  sin'^/i  —  cos/u.  —  (1  —  cos /a)  (1  —  c"*  sin"''/u,)*  ^ 
siu'^/i  ' 

therefore,  if  c'  be  the  distance  from  its  centre  to  its  focus, 

,a  _  1  —  cos  yu,  —  (1  —  cos /a)  (1  —  (?  sin^/i)* 

C       —    ;: — o 

sm  fx 

1  —  cos/A  -  c'"^  sin'''/u,  —  (1  —  cosyu,)  (1  —  c^sin^/i)* 
sin'"' /A 

•••       C'   =  «5 

whence  the  ellipses  are  confocal. 

14.  If  cc,  y^  z,  r«,  are  linear  functions  of  the  coordinates  of 
any  point,  such  that  no  three  of  the  lines  represented  by  the 
equations  «  =  0,  2/  =  0,  z  =  0,  ?i;  =  0,  meet  in  a  point,  the 
equation  w  +  [yzf  +  [zx]^  +  [xy)^  =  0  is  that  of  a  cui've  of  the 
fourth  order  having  three  double  tangents,  x  =  0,  y  =  0,  z  =  Oj 
and  three  double  points,  y  =  z  =  w,  z  —  x  =  w^x=y  =  w. 
Shew  also  that  the  six  points  of  contact  of  the  double  tangents 
lie  in  a  conic  section. 

Where  the  line  a?  =  0  meets  the  cun'^e, 

w  +  [yzY-  +  {zx)^  +  {xy)^  =  0 (1); 

we  have  also     w  +  [yz]^  =  0,     or  w''  =  yz , (2). 


1851. J  GEOMETRY   OF  TWO   DIMENSIONS.  163 

It  hence  appears  tliat  the  line  x  =  0  meets  (1)  only  in  the 
points  where  it  meets  the  conic  (2),  that  is,  in  two  points  only. 
But  (1)  when  rationalized  takes  the  form 

(«<?'■'  —  yz  —  zx  —  xyY  =  ^^xyz  [x-Vy-k-z-  Iw) (3), 

mider  which  form  we  see  that  the  curve  represented  by  it  is 
of  the  fourth  order;  therefore  the  line  a;  =  0  must  meet  it  in 
four  real,  coincident,  or  imaginary  points.  Therefore,  cither 
each  of  the  pomts  in  which  a-  =  0  meets  (1)  must  be  a  double 
point,  or  a;  =  0  must  be  a  double  tangent  to  (1) ;  for  from  con- 
siderations of  symmetry  it  is  clear  that  both  points  must  be  of 
the  same  nature.  Also  we  see  that  ^  =  0,  2  =  0  stand  in  ex- 
actly the  same  relation  to  (1)  as  a?  =  0  does. 
Now  consider  the  conic  whose  equation  is 

xc^  —  yz  —  zx  —  xy  =  0  (4) . 

From  (I),  (2)  we  see  that  all  the  points  in  which  x  =  0, 
y  =  0,  s  =  0,  meet  (1)  lie  in  this  conic.  Hence  these  points 
cannot  be  double  points,  for  if  they  were,  a  cm've  of  the  second 
order  would  intersect  a  cm've  of  the  fourth  order  in  twelve 
points  (coinciding  two  by  two),  which  is  impossible.  Therefore 
X  =  0^  y  =  0,  2  =  0,  are  double  tangents  to  (1). 

Again,  if  in  (3)  we  put  x  =  u',  it  becomes 

{x  [x  —  y  —  z]  —  yz\-  =  4.xyz  {y  +  z  —  x), 

which  can  be  reduced  to 

X  [x-y-z]  +yz  =  0 (5), 

shewing  that  the  line  x  =  w  meets  the  given  curve  only  in 
the  points  in  which  it  meets  the  conic  (5),  that  is  in  two  points 
only.  Hence,  either  x  =  w  is  a  double  tangent,  or  it  must  meet 
the  cun'c  in  two  double  points. 

Now  at  the  points  where  it  meets  the  curve,  we  have,  as 
may  be  seen  from  (5),  z  =  x  =  ?r,  x  =  y  =  lo  respectively. 

Hence,  where  the  line  x  =  tv  meets  the  cm'vc,  it  also  meets 
either  the  line  y  =  lo  or  z  =  iv^  and  from  considerations  of 
symmetry,  if  x  =  to  touch  the  ciu-ve,  y  =  iv  and  z  =  w  muM 
do  so  likewise  at  the  same  points. 

m2 


1()4  SOLUTIONS    OF    SKNATi:- HOUSE    PROBLEMS.  [1851. 

Now  we  have  already  shewn  that  if  x  =  w  do  not  touch  the 
curve  (I),  it  must  meet  in  two  double  points,  and  we  have  just 
now  proved  that  if  it  does  touch  the  curve,  y  =  lo  and  z  =  w 
touch  it,  each  at  one  of  the  points  where  x  =  w  docs.  Therefore 
two  tangents  can  be  drawoi  at  these  points  in  different  direction?, 
therefore  they  must  be  double  points.  Hence,  in  either  case, 
the  points  in  question  are  double  points,  viz.  z  =  x  =  to  and 
X  =  y  =  w\  and  similarly,  it  may  be  shewn  that  y  =  z  =  w  \9 
a  double  point. 

15.  (a).  Describe  a  circle  when  two  tangents  are  given, 
and  a  point  from  which  a  pair  of  tangents  drawn  to  the  circle 
shall  include  a  given  angle. 

[^).  By  means  of  the  properties  of  reciprocal  polars,  or 
otherwise,  construct  a  conic  section,  when  the  focus,  two  points, 
and  the  angle  between  the  asymptotes  are  given. 

(a).  Let  AB^  AC  (fig.  77)  be  the  two  tangents,  P  the  point 
from  which  a  pair  of  tangents  are  to  include  a  given  angle  a.  Let 
lBA  C  =  13.  Bisect  the  angle  BA  C  by  the  straight  hue  ^D, 
join  PA^  and  divide  it  in  E^  so  that  PE :  EA  : :  sin  ^/3  :  sin  ^a. 
Also  produce  PA  to  F^  so  that  PF:  ^i^: :  sin^/S  :  sin  ^a. 
Bisect  EF  in  O^  and  with  G  as  centre  and  GE  as  radius, 
describe  a  circle,  cutting  AD  in  H.  H  shall  be  the  centre  of 
the  required  circle. 

For  since  PE :  EA  : :  PF :  FA  : :  sin^/3  :  sin  ^a,  the  locus 
of  a  point,  the  ratio  of  whose  distances  from  P  and  A 
=  sln^/3  :  sin^a  is  the  circle  of  which  EF  is  a  diameter. 
Therefore,  joming  PH,  PH :  AHw  sin^/3  :  sinia. 

Draw  PK  a  tangent  to  the  circle  whose  centre  is  H  and 
which  touches  AB^  A  C,  then  if  r  be  the  radius  of  that  circle, 

-^  =  sin  EPK,      -^  =  sin  1/3 ; 

.-.    sin  HPK  :  sin  ^/3::AH:PII 

: :  sin^a  :  sin  ^/3j 


HPK 


a. 


2'^1 


ISoL]  QEOMETltY    OF   TWO    DIMENSIONS.  165 

and  a  pair  of  tangents  to  the  circle  drawn  from  P  include  the 
required  angle  a. 

{13).  If  we  take  the  polar  reciprocal  of  the  above  system 
with  respect  to  P,  to  the  circle  will  correspond  an  hyperbola 
whose  focus  is  P;  to  the  two  given  tangents  AB^  AG  coitc- 
spond  two  given  points,  and  to  the  tangent  through  P,  including 
a  given  angle,  correspond  two  points  on  the  cm'vc  at  an  infinite 
distance  subtending  a  given  angle  at  the  focus  or  (since  the 
points  arc  infinitely  distant)  at  the  centre.  This  angle  therefore 
is  the  angle  between  the  asymptotes.  Kence  (/8)  is  the  polar 
reciprocal  of  (a);  and  therefore  the  requii'ed  conic  section  may 
be  constructed  by  means  of  the  circle  there  determined. 

16.  Let  P  be  any  point  in  a  conic  section  whose  focus  is  8 
and  eccentricity  e ;  in  >SPtake  SQ  =  \L  (the  semi-latus-rectum); 
draw  QR^  ST  perpendicular  to  ;SP,  meeting  the  tangent  at  F 
in  R  and  T  respectively;  also  draw  SY  perpendicular  to  the 
tangent  meeting  it  in  1^;  and  let  PZ7,  QZ  drawn  parallel  to  the 
transverse  axis  meet  8Y  in  U  and  Z  respectively :  it  is  required 
to  prove  one  of  the  following  properties : 

(1)  P  is  a  point  in  the  latus-rectum. 

(2)  QR  passes  through  the  point  U. 

(3)  PU=e.PS.         (4)   SR  =  e.ST.  (5)  SY.SZ={^L)\ 

(1).  Let  the  inclination  of  SP  to  the  axis-major  be  a,  then 
the  polar  equation  to  the  tangent  at  P  will  be 

1       2 

-  =  -^{(cosi9  +  cos(^-a)}; 

and  that  to  QR,  r  =  \L  sec  (^  —  a). 

At  P,  the  point  of  intersection  of  these  lines,  we  must  have 

^  =  |7r; 
therefore  R  is  a  point  in  the  latus-rectum. 

(2).    The  equation  to  /ST  Is 

/,  sin  a 

tan  0  =  , 

e  +  cosa 


16G  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1851. 

or  ill  rectangular  coordinates, 

jc  ama  =  y  {e  +  cos  a). 

Now  the  ordinate  of  P  is 

i^L  sin  a 
1  +  e  cos  a  ' 

therefore  the  equation  to  PU  is 

^L  sin  a 


y  = 


1  +  e  cosa 

At  U  the  point  of  intersection  of  these,  we  have 

_  ^L  (e  4- cosa) 
1  +  e  cosa    ' 

.'.   X  cosa  +  y  sin  a  =  ^L. 

Now  this  is  the  rectangular  equation  to  QB ;  hence  QR  passes 
through  the  point  U. 

,  V     m,       1     •         n  r^-     \L  cosa 

(3).    The  abscissa  of  P  is . 

^  '  1  +  e  cosa 

That  of  CThas  been  shewn  to  be 

^L  (e-fcosa) 
1  +  e  cosa    ' 

and  P,  ?7,  have  the  same  ordinate ;  hence 

PU==e.PS. 

(4).  Since  i2  is  a  point  in  the  latus-rectum,  SR  is  perpen- 
dicular to  PZ7,  and  ST  is  perpendicular  to  6T,  Pii  to  SY^ 
whence  it  readily  follows  that  the   triangles   STR,   SPU  are 

similar ; 

.-.    PU'.PS::  SR:  ST', 

but   PU=e.PS, 

.:    SR  =  €.ST. 

(5).    The  polar  equation  to  QZ  is 

r  sin  ^  =  ^L  sin  a, 


1851.]  GEOMETRY   OF  TWO   DIMENSIONS.  167 

and  that  to  ^F  is 

zj  sin  a 

e  +  cosa 

.    ^  sina 

.*.   sma  = 


;H-2e  cosa  +  e')*' 
therefore  at  Z^  the  intersection  of  these  lines, 

r  =  6'Z  =  ^X  (1  +  2c  cosa  +  e')*. 

Also  SY=^\L  ^ 


{(e  +  cosa)'^  +  sin*  a}* 

1 ^ 

(1+26  cosa  4- e')*' 
.-.    8Y.8Z=[\L)\ 


==^L 


17.  If -4j,  A^...A^/y  rtj,  «2---'^i5  ^^  ^^^^  angular  points  of  two 
polygons  of  n  sides  each,  which  circumscribe  a  given  circle,  and 
Pj,  P^...F^  the  points  of  intersection  of  their  first,  second... h"' 
sides  respectively ;  shew  that 

Shew  also,  by  means  of  projective  properties  or  otherwise,  that 
the  same  equation  is  true  when  any  conic  section  is  substituted 
for  a  circle. 

From  0,  the  centre  of  the  circle,  draw  perpendiculars  Oi?,, 
OB^...OB,^,  0\,  0\...0h^,,  on  the  sides  A^A^,  A^A^...A^A^, 
a,a^^  ^,«2*'-^n-i^«?  respectively.  Through  0  draw  any  line  OX, 
and  let  generally  B,OX  =  a„  h^OX  =  /3,  (fig.  78).     Then 

A^OX  =  i (a,_,  +  aj,     .-.   AOB,  =  A,OB,_^  =  ^ (a,  -  a,J. 
Similarly,  a,  OB,  =  M^r  -  A-J, 

and   P,C>Z=i(a, +  ^,); 
.-.   P,0j5,  =  P^OX  -  B^OX  =h{0r-  aJ- 
Also  PA,  =  a  (tan ^. OP.  +  tan P, OA" ) , 

a  being  the  radius  of  the  circle, 

=  a{tan^(a^  -  a,_J  +  tan^(|S,  -  a,.)| 

_^^  sin^(/3,  -  a,_J 

cos-^  (a^  -  a^_,)  cos  ^  (/3^  -  aj  * 


168  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1851. 

Also  F.l      =  a  (tan  yl,.,,  OB,.  -  tan  P,  OB) 

=  a  {tan^  (a,.,,  -  a,)  -  tan^  (/3,  -  a,,)} 
=  a  sin^(a,^,  -^,) 

The  expressions  for  P^a,.  and  P^a^+j,  will  be  got  from  these 
by  simply  interchanging  a  and  y9 ; 

.     Pr,  -n  sin^(a. -/3,_J 

•  •    ^^''^  -  ""  eosi  (^,  -  /3,_J  cosi  (a,  -  ^J  ' 


Hence 


cos^(/?,.^j  -  ySJ  cosi(a,  -  ^,. 
P.^...P.a. 


PA..^,.Pa 


^  sin  I  (^,  -  a,_,)  sin  ^  (a,  -  ^,._, )  cos  |  («,^^  -  g,.)  cos  ^  (^,.^,  -  ^  J  , 

sin^(^,,^j-a,.)sin^(a,.^j-/3,.)cos-^(a,-a,_,)cos^(^,.-^,_J 

From  the  form  of  this  expression  it  is  easy  to  see,  that  if  we 
give  r  every  value  from  1  to  n  inclusive  and  multiply  the  re- 
sulting fractions  together  (observing  that  instead  of  a,.^j,  /3,.^,,  we 
write  otj/SJ,  every  factor  will  appear  both  in  the  numerator  and 
denominator.     Hence 

P,^,.P,a,.P,^,.P,a,...PA-i^A  _  1 
PA,P.a,^PAs'^.%'"KA-PA         ' 
or   P^A^.P^a^.P^A,^.P^a^. .  .P^.^^A 

jf  f       T>  i  1   .-.  .     OP,..OA,..smP,.OA, 

It   tor   P,.A,    we   substitute ^^^ -,  and  make 

similar  substitution  for  each  of  the  other  lines,  each  member  of 
this  equation  will,  since  OB^  =  a,  be  divisible  by 

OA,.  Oa,.  0P\ . .  0A„.  Oa.  op: 


and  there  will   remain  merely  a  relation  between  the   sines  of 
angles  subtended  at  0.     The  property  just  proved  must  there- 


1851.]  GEOMETRY    OF   TWO    DIMENSIONS.  169 

fore  be  tnie  for  any  figure  into  which  the  circle  can  be  projected, 
that  is  for  any  conic  section.  (See  the  article  on  the  Method  of 
Projections,  in  Salmon's  Conic  Sections^  Chap.  XIV.) 

18.   Prove  one  of  the  two  follo%viiig  properties. 

(1).  When  one  of  the  foci  of  a  conic  section  and  two  tan- 
gents are  given,  the  locus  of  the  other  focus  is  a  straight  line. 

(2).  When  the  centre  of  the  conic  section  and  two  tangents 
are  given,  the  locus  of  the  focus  is  an  equilateral  hyperbola. 

The  proof  of  these  theorems  depends  on  the  property,  that 
the  product  of  the  pei'pendiculars  from  the  foci  on  the  tangent 
at  any  point  of  a  conic  section  is  constant  and  equal  to  the 
square  of  the  semiaxis  minor. 

(1).  Let  J.P,  AQ  (fig.  79)  be  the  two  given  tangents,  S  the 
given  focus,  H  that  whose  locus  is  to  be  foimd.  Draw  /S'F,  HZ 
perpendicular  to  AP\  SY\  HZ'  io  AQ'^  then 

SY.HZ=  ST. HZ', 

.-.   HZ'.HZ::  ST  :  SY, 

a  constant  ratio ;  therefore  the  locus  of  His  a  straight  line. 

(2).  Take  the  centre  as  origin,  and  let  the  equations  to  the 
given  tangents  be 

cc cosa  +  y  s'moL  —  a  =  0  (1),     a^cosa'  +  i/  sina'  —  «'  =  0  (2). 

Let  I,  77,  be  the  coordinates  of  one  focus,  then  —  ^,  —  v^  will 
be  those  of  the  other.  Now  the  length  of  the  perpendicular 
from  I,  7],  to  (1)  is 

^  cosa  +  T)  sina  —  a; 

similarly,  that  from  —  ^,  —  77,  is 

—  I  cosa  —  77  sina  —  a. 
Hence  we  get 

(I  cosa  +  77  sina)'"'  -  n'  =  ^\ 

/3  being  the  semiaxis  minor. 

Similarly  it  may  be  shewn  that 

(^cosa'  -f-  77  sina'V^  —  «'^  =  ^'-^ 


170  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

.•.    (^  cosa  +  7}  sina)"''  —  a"  =  (|  cosa'  +  rj  siiia')"'  —  «'^ 

or   (I  cosa  +  r)  sina)'^  —  (^  cosa'  +  r/  sina')"''  =  «■*  —  « '^ 

the  equation  to  the  locus  of  |,  t;,  which  is  therefore  a  rectangular 
hyperbola,  the  equations  to  whose  asymptotes  are 

I  (cos  a  +  cos  a')  +  t;  (sin  a  +  sin  a')  =  0, 
f  (cosa  — cosa')  +  ■j?  (sina  — sina')  =  0. 


171 


DIFFERENTIAL  CALCULUS. 

1848. 

Find  the  equation  to  that  involute  of  a  cycloid  which  passes 
through  the  cusp,  and  shew  that  in  the  immediate  neighl)om'hood 
of  the  cusp  it  becomes  the  curve  2a{4:yY  =  {3x)*y  a  being  the 
radius  of  the  generating  circle. 

Let  X,  y  be  the  coordinates  of  any  point  in  the  cycloid 
referred  to  the  cusp  as  origm,  and  base  as  axis  of  ic,  s  its  dis- 
tance measured  along  the  arc  from  the  cusp ;  ^,  rj  those  of  the 
corresponding  point  in  the  involute.  The  equation  to  the 
cycloid  will  be 

X  =  a(^  — sin^), 

y  =  a[\  —  cos^), 
and  we  have,  since  the  tangent  at  {pcy)  passes  through  (f?;)  at 
a  distance  s  from  [xy]^ 

^  dx 

dy 

dx 

Now  32  =  rt(l  —  cos^)  =  2asiii''^^, 

do 

-^=      asin^       =  2a8ini^  cos^; 
da 

dd  ^   ' 

and  s  =  4a(l  -  cos|^) ; 
.-.  f  =  a(^-8in6')  -  4a(l-co8^^)  sin^^, 
=  a(^  +  sin^-4sm^^), 
7)  =  a{l  —  cos^)  —  4a (I  —  cos^^)  cos^^, 
=  «(3  +  cos^  -   Icos^^), 
the  equations  to  the  involute. 


172  SOLUTIONS    OF   SENATE-HOUSK    PROBLEMS.  [1849. 

In  the  iiiinicdiatc  neighbourhood  of  the  origin  where  6  is 
small,  these  become 


and  77  =  a  j  4  — - 

6'       . 
+  24-^ 

f-.f; 

6" 
=  %.4- 

Hence,  eliminating  6^ 

a'-- 

=(f)' 

or 
the  required  curve. 

■  2«(477)^ 

=  m% 

1849. 

1.  K  P  be  a  point  in  a  cycloid,  and  0  the  con-esponding 
position  of  the  centre  of  the  generating  circle,  shew  that  PO 
touches  another  cycloid  of  half  the  dimensions. 

Let  a(^  — sin^),  a(l— cos^)  be  the  coordinates  of  P,  as  in 
the  last  problem ;  then  a6  and  a  will  be  those  of  0. 
The  equation  to  PO  is 

X  —  aO      y  —  a 
sm  V  cos  V 

or  X  cos Q  -^  y  ^VQ.Q  —  a{6 cos Q  +  sin 6). 

Differentiating  this  equation  with  respect   to  Q  as  variable 

parameter, 

—  a;  sin  ^  +  ?/  cos  ^  =  a  (2  cos  Q  —  ^  sin  &) ; 

.*.  a;  =  a(^  — sin^  cos^), 

=  ia(26'-sin2^) (1), 

and  ?/  —  r;  ( 1  +  cos'*  &) 

=  \a\\  +cos^^) (2). 


1849.]  DIFFERENTIAL  CALCULUS.  173 

Equations  (1)  and  (2)  shew  that  the  line  OP  always  touches 
a  cycloid  whose  cusp  coincides  with  the  cusp  of  tlie  original 
cycloid,  and  generated  by  a  circle  of  half  the  size  of  its  gene- 
rating circle. 

2.  Find  the  locus  of  the  ultimate  intersections  of  the  lines 
defined  by  the  equation 

a;cos3^  +  ?/sin3^  =  a(cos2^)* (1), 

where  6  is  the  variable  parameter. 

Differentiating  (1)  with  respect  to  ^, 

a;sin3^  -  ?/cos3^  =  asin2^  (cos^)* (2). 

Squaring  (1)  and  (2),  and  adding, 

x'  J^  f  =  d'cos2d (3). 

Again,  (2)  -i-  (1)  gives 

ccsin3^  —  ?/cos3^      ,      ^^ 

— z — ^    .    ,^  =  tan  2^, 

a;cos3c7  +  ysm'do 

or  — ^— — — —  =  tan  2^; 
1  +  tan3^^ 

X 

.-.  ^=  tan^, 

X 

1  1      /«\     -1        -1        2  1  —  tan*'^ 
and  by  (3)  x'  -\-  i/  =  a'  - 


=  a 


+  tan' 
x'^-f 


d'  +  f 

or  {x^  +  yr-^^\^'-f)^ 
the  equation  to  Bcmouilli's  Lemniscate. 

3.  If  e  be  the  eccentricity  of  a  conic  section,  r  the  distance 
of  any  point  from  the  focus,  p  the  radius  of  curvature  at  that 
point,  and  ds  an  element  of  the  arc  of  the  curve,  then 

,      Jr'        ,  fd'r 

'  =d?-^PW' 


174  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

Let  the  equation  to  the  coiiie  section  referred  to  its  focus  as 
origin  and  axis-major  as  axis  of  a?,  be 

x'  +  f  =  [ex  +  cY ; 
.*.  r  =  rx  +  c, 

dr 
,  dr  _d'C  _  e 


ds      ds       J         ('hf\ 
dx      \         \  dxj 
d'^r        d   dr     I 
ds'       dx  ds  '  ds  ' 
dx 

e^  it 
dx''  dx         1 


-||)T'I' 


dy^ 

e  dx 


"^-(IJ 


drV         ,  fdS'Y        ,  ^  +  \dx) 


dlX 


ds)  ^P  Uv  ~''        7±V' 

"^  [dxJ 
=  e\ 

4.   If  u  be  a  function  of  the  independent  variables  x,  ;/,  z^ 
given  by  the  equations 

^*=/(^,  0 (1), 

s  =  F{Ix  +  w??/  +  nz  +  ht) 
=  0  [mz  —  ny)  +  % [nx^  ~Iz)  +  yjr  [ly  —  mx)^ 
and  if  V  +  m'  +  n'  =  F ;  shew  that 

,  du  du         du       ,  du 

aa;  dy  dz  dt 

where  -j-  is  obtained  from  (1)  by  considering  s  constant. 


1850.]  DIFFERENTIAL   CALCULUS.  176 

We  have 

also  =z  n')(^  —  wi/r' ; 

dt        ??^'  —  7>li|r'         I 

similarly    |  =  |/(.)  +-0)|  (j-^^'  -  «f )  -  |/((), 

Multiplying  these  equations  by  ?,  wi,  w  respectively,  and 
adding,  remembering  that  1^  +  m"  +  w"  =  ^'^j 

,^M  c7m         f?M      ,  du      . 

^  J-  +  ?w  -J-  +  w  -J-  +  «  -^  =  0, 
GKC  dy         dz  dt         ' 

du  „,,  , 

smce  -J  means/  [t). 

1850. 

1.  A  paraboloid  of  revolution  with  its  axis  vertical  contains 
a  quantity  of  water,  into  which  is  sunk  a  heavy  sphere,  and  the 
water  is  just  sufficient  to  cover  the  sphere ;  find  the  form  of  the 
paraboloid  that  the  quantity  of  water  with  which  this  can  bo 
done  may  be  the  least  possible. 

Let  a  be  the  radius  of  the  sphere,  I  the  latus-rectum  of  the 
paraboloid;  h  the  height  to  which  the  water  rises  when  the 
sphere  is  siuik :  then  if  C  be  the  content  of  the  paraboloid  of 
height  ^,  V  the  volume  of  the  sphere,  Q  the  quantity  of  water, 

g  =  o-r; 

and   we   have   to   make    Q    a   minimum   by   the   variation    of 
I  and  h. 


176  SOLUTIONS   OF  SENATE-HOUSE   PK0BLEM8.  [1850. 

Now  C=l'rrlh', 

therefore  IK^  must  be  a  minimum. 

Now   from  the  vertex  the  equation  to  the   section  of  the 
paraboloid  is 

/  =  ^x ; 

that  to  the  section  of  the  sphere  is 

[x-  {h-a)Y  +  y'  =  d\ 

In  order  that  these  may  touch  one  another,  we  must  have 

[x  —  {k  —  a)Y  +  Ix  —  a\ 

a  perfect  square,  which  requires  that 

4:{h'-2al)  =  {l-2{h-a)]% 

or    r  -  U[h-a)  +  4a'  =  0  (l). 

Hence  we  must  make  JK''  a  minimum  subject  to  the   con- 
dition (1),  which  may  be  written 

£+_2ar. 

4Z        ' 
therefore  we  have  to  make 


(Z+2aV 

=  mmmium 


I 


4  1 


or  ^ T  =  0 ; 

.-.    ?  =  fa; 
which  determines  the  form  of  the  paraboloid. 

2.  If  a  circle  be  described  touching  a  curve  at  any  point 
(r,  6)  and  passing  through  the  pole,  shew  that  the  equation  to 
the  circle  will  be 

do  r 

The  general  equation  to  a  circle  passing  through  the  pole 
and  the  point  (r,  6)  is 

r  —  r  sec(^—  a)  cos(^'  — a)  (1), 


1850.]  DIFFERENTIAL   CALCULUS.  177 

a  being  the  angular  coordinate  of  the  diameter  through  the  pole. 
This  equation  may  be  put  in  the  form 

cos{6'-a  + (^'-^)| 
j«  ^=  )• '■      !^ ~ 

cos  [d  —  a) 
=  r  {cos  {&  -6)  -  tan  {6  -  a.)  sin  {&  -6)]. 

Now  from  (1),     'ia>  =  ~  '''  ^^^  (^  ~  '^)  '^'^  (^  ~  ^) 

=  -j^  ,  when  6'  =  6^ 
do 

since  the  circle  touches  the  curve  at  the  point  (?•,  6)\ 

•'■    *■  r  tan  (^  -  «)  =  ^  ) 
and  r  =  r  {cos  [6'  -  ^)  +  "4  sin  [6'  -  6)] 


,  d  sm[e'-e) 


~       '*  dd  r  ' 

,  d    sm{0'-0)       ^ 

or     r'  +  r'  ^   ^ '  =  0. 

dU  r 

3.  If  a  parabola  roll  upon  a  line,  the  focus  will  trace  out 
a  catenary. 

The  following  more  general  problem  admits  of  very  easy 
solution:  A  given  curve  rolls  upon  a  straight  line,  to  find  the 
locus  of  any  point  to  which  the  curve  is  referred  as  pole. 

Let  AB  (fig.  79)  be  the  given  straight  line,  A  any  fixed 
point  in  it.  Let  CF  be  the  rolling  cm've,  C  the  point  which  has 
been  in  contact  with  A,  S  the  pole,  P  the  point  of  contact  in  the 
position  represented  in  the  figure.  Join  ^SP  and  di'aw  /SF  per- 
pendicular to  AB.  Let  A  Y  =  x,  YS  =  y,  SF  =  r,  s  the  arc 
of  the  cun^e  described  by  S.  Then  the  tangent  being  manifestly 
pei*pendicular  to  SF,  we  have 

'^  =  cohFSY=^, 
ds  r 

*  For  tills  solution,  wc  are  indebted  to  Mr.  Goodwin 

N 


178  H(.>LUTl(>NS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

Let  the  equation  to  the  rolling  curve  be  r''  =f{p)i  then  the 
equation  of  the  required  locus  is 

^■'{•H- (I)}  =/(.). 

In  the  case  of  the  parabola,  we  have 

'■=?^ 

dx      y       c 
ds       r       yl 

the  differential  equation  to  the  catenary. 

4.    Find  the  forms  of  the  curve  whose  equation  is 


,ii  I  ^i- 


xy  =  m  [x  +  y  —  «  ], 
according  as  nt^  is  >  =  or  <  -^  a^. 

Arranging  the  equation  as  a  quadratic  in  cc,  we  have 

and  a^  =  ^  +  A  {/  +  4/n'  («'-/)}*  (1). 

Hence  we   may  use  x  =  -^.^  as  a  guiding  cm've ;  its  fonn 
is  shewn  by  the  dotted  curve. 
To  consider  the  equation 

or  f  -  4w* .  f  +  ^m\t'  =  0. 

This  equation,  considered  as  a  cubic  in  y\  will  have  three 

3  3J 
real   roots  or  one,   according  as   oti'  is  >    or   <   — —  a'"*.*    If  it 

has  three  real  roots,  one  of  them  is  negative,  and  the  corre- 

3  3- 
*  If  m  =  -j—  «^,  it  vdH  have  three  real  roots,  tAvo  of  them  being  equal. 


1850.J  DIFFERENTIAL   CALCULUS.  179 

spending  values  of  y  imaginaiy.  If  it  has  two  equal,  and  there- 
fore two  real  roots,  the  equal  roots  are  positive  and  the  other 
negative,  giving  only  one  positive  value  to  y'.  If  it  has 
only  one  real  root,  it  is  negative  and  y  imaginary.  Again, 
differentiating  the  original  equation,  we  get 

dy         2m'x  —  y^ 


dx      ^fx  -  2m'y  ' 

therefore 

1,  when 

y 

=  0, 

and  therefore  x  =  a^ 
dx 

Also, 

taking 

the  lower  sign  in  equation  (1), 

X 

= 

2m' 

2  m    \                     y^ 

-1 

= 

—  m 

2                2 

'"V       &c. 

=  0  if  ?/  =  GO  : 

hence  the  axis  of  y  is  an  asymptote  to  the  cui-ve,  which  we 

thus  see  has  the  fonn  represented  in  figs.  80,  81,  82,  according 

3.3- 
as  m'  is  >  =  or  <  ~ —  a^. 
4 

5.    Trace  the  curve  whose  equation  is 

2a 
[r  —  cf  =  cd  {2a  —  cO)  when  c  =  —  , 

TT 

and  prove  that  as  c  increases  indefinitely  the  cui've  approximates 
to  a  circle. 

When  6  is  positive,  c6  must  be  less  than  2a  or  6  loss  than 
TT,  and  6  can  receive  no  negative  value.  Also,  any  value  6^  of  0 
gives  the  same  value  for  ?•  as  tt  —  6^.      Solving  the  equation 

we  have 

r  =  c±  [cd[2a-ce)]^ 


2a 

=  —  +  2a 


K(-^)r <■)• 


N  2 


180  SOLUTIONS   OF   SENATE-HOUSE    PHOBLEMS.  [1850. 

The  quantity  affected  with  the  ambiguity  lias  its  greatest 
a 
value  when  -  =  ^,  when  r  receives  the  value 


2a 
r  =  —  ±  a 

IT 


the  latter  of  which  is  negative.     Again,  putting  ^  =  0,  we  have 
r  =  c]  and  differentiating, 

therefore  when  r  =  c,  and  therefore  ^  =  0, 

dr 

33  =  °"- 

Hence  the  cui've  is  of  the  form  shewn  in  fig.  83. 

Also,  as  c  and  therefore  a  is  indefinitely  enlarged,  equation  (1) 

becomes  >•  =  —  ,  representing  a  circle. 

TT 

6.  Find  the  locus  of  the  consecutive  intersections  of  the  curve 
whose  equation  is  x^  +  y'^  =  2ax'  +  2by'  (1) ;  a  and  h  having 
any  values  which  satisfy  the  equations 

2„  (.!-,)  =  (."-/)  I -2., (2), 

2i(4-,)=.'-y'  +  2.,| (3), 

X  and  y  being  the  coordmates  of  any  given  curve. 

The  problem  is  best  solved  by  introducing  polar  coordinates. 
Let  X  =  r  cos  6^     y  =  r  smO] 

.-.  dx  =  —  r  &mdd6  -+  cosOdr^     dy  =  r  co^6dd  +  shiddt", 

.'.  xdy  —  ydx  =  r^dd^ 

[x^  —  y^)  dy  —  2xydx  =  r^  cos2ddy  —  r^  sm20dx 

=  r^  cos  Odd  —  r^  s,m  6 dr^ 
{x^  —y'^)  dx  +  2xydy  =  r^  sin  Odd  +  r^  cos  6 dr. 


1851.]  DIFFERENTIAL  CALCULUS.  181 

Hence  equations  (2)  and  (3)  become 

2a  =  r  cos  6  —  sin  6  -f^ , 
da 

2b  =  ?•  sin  6  +  cos  0  -j^  : 
au 

da  .     .  ^'r 

.-.  2^=->-sin^-sm^^, 

2  -77:  =  r  cos  c?  —  cos  U  ^7^  . 

Also  equation  (1)  transformed  into  polar  coordinates  becomes 
r'  =  '2a  cos  0'  +  2b  sin  0'. 

Differentiating   this   equation    with    respect    to    0,    considering 
r'  and  0'  constant,  we  have 

0  =  2  -77.  cos^  +2-77=  sm^ 
dtf  dv 

=  r&m{0'-0)-cos{0'-0)'^,- 


tan  (^'-6')  = 


dff' 


From  this  equation,  when  r  has  been  substituted  in  terms 
of   0^  from  the  known   equation   to  the   curve,  we  can  find  0 

in  tei-ms  of  0'',  and  thence  r  and  -^  will  be  known  in  tcnns 

dtf 

of  0'j  and  the  required  equation  to  the  curve  will  be 

r  =  2a  cos^'  +  2b  sin^' 

dr 


=  rcos{0'-0)  +  Hm{0'-0)%. 

d0 


1851. 

If  (f)  (c)  be  a  rational  and  integral  function  of  c,  the  coeffi- 
cients of  which  are  functions  of  any  number  of  variables  ^,  ?/,,.. 
then  if  8  denote  differentiation  with  respect  to  the  variables, 
and  the  quantity  c  be  eliminated  from  the  equations  0  [c]  =  0, 


182  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

3(/)(t')=(),  the  result  may  be  represented  by  tiSPSQ...  =  0 
where  P,  Qy..  are  the  roots  of  the  equation  <f>{c)  =  0,  and  tc  =  0 
is  the  result  of  the  elimination  of  c  from  the  equations  (p  (c)  =  0, 
<f>'{c)=0. 

Since    0(c)  is  a  rational  and   integral   function   of  c,  and 
P,  Qy..  are  the  roots  of  the  equation  0(c)  =  0,  we  have 
0  (c)  =  [c  —  P){c—  Q) identically. 

dP  dP 

Let  P^  denote  ^-  dx.    P.  -7-  diu...  then 
^  c?^      '       "'  dy    '^' 

g0(c)  =  0(c)  Slog  0(c) 

=-*w(^^^^-^^--) 

=-{(^.+^.+-)(o-^)(c-^)-+(^.+^.+..-)(c-^)(«-^)+"l- 

Hence  the  result  of  the  elimination  of  c  between  0(c)  =  0 
and  S0(c)  =  0,  is 
0  =  product  of  the  expressions  (P^.  +  P^+...)  (P-  Q)  (P-  /?)..., 

((?.+  (?,+•••)  («-P)(^-P)...,&c. 

=  (P,+P^+...)(^,+  ^,+...)...(P-(?)(P-P)...(^-P)((2-P)... 

=  v  suppose. 

Again,    0'(c)  =  (c-^)  (c-P)...+  (c-P)  (c- P). ..+...  ; 

.-.    ,,=  (P-^)(P-P)...(^-P)(^-P)..., 

and    SP=P,  +  P, +...,     8^=  ^,.+  ^^4-...; 

.-.    v  =  uBPBQ.... 

Hence  the  result  of  the  elimination  may  be  represented  by 
u8P8Q...  =0. 


(  l«-i  ) 


INTEGRAL  CALCULUS. 

1848. 

The  comer  of  a  sheet  of  paper  is  tunied  down  so  that  the 
sum  of  the  edges  turned  down  is  constant ;  find  the  equation  to 
the  curve  traced  out  by  the  vertex  of  the  angle ;  find  also  the 
area  of  the  curve. 

Let  r,  ^,  be  the  polar  coordinates  of  the  vertex,  refeiTed  to 
the  origmal  position  of  the  vertex  as  pole,  then  the  lengths 
of  the  respective  edges  are 

^r  sec  0,     ^  r  cosec  ^,     respectively  ; 
therefore  the  equation  to  the  curve,  is 

l^?- (sec  ^  + cosec  ^)  =  constant  =  a  suppose, 
or  in  rectangular  coordinates, 

{x  +  7/)  {x'  +  f)  =  2axy. 
To  find  the  area,  turn  the  axes  through  an  angle  ^tt,  then 
we  get  _acos26'_ 

*"  ~  2icos^  ' 
therefore  if  A  be  the  area  of  the  loop  traced  out  by  the  vertex, 

A=-l        rW 


1  r\^ 


i-^  cos' 2^   ,^ 

27j     ff" 

COS  a 


4cos''^-4+-^]  iW 
cos  a/ 


2cos2^-2  +  6ec'<9)<Z^ 


=  |(2-7r  +  2) 


184  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1848. 

2.  Tangents  to  a  system  of  similar  and  concentric  ellipses 
arc  drawn  at  a  given  pei*pendienlar  distance  from  the  centre ; 
find  the  locus  of  the  point  of  contact,  and  shew  that  the  area 
of  the  curve  is  equal  to  that  of  an  ellipse  which  has  the  same 
greatest  and  least  diameters. 

Take  the  common  axes  of  the  ellipses  as  coordinate  axes, 
and  let  c  be  the  distance  of  each  tangent  from  the  centre,  then 
if  6  be  the  inclination  of  a  pei-pendicular  on  any  tangent  from 
the  centre  to  the  axis  of  a;,  the  equation  to  that  tangent  will  be 

cccos^  +  ymiO  =  c (1). 

Let  the  equation  to  any  one  of  the  ellipses  be 

2  'i 

7?  +  !?  =  '"  C^)- 

If  (^,  7))  be  the  coordinates  of  the  point  where  (1)  touches 

this,  we  have 

f         cos^         7?        sin^ 


'     ...n:^         ^     ' 


and  since  |,  r;,  is  a  point  in  (2), 

m'  =  K  +  t' 
a:       h 

Eliminating  m^  6  between  these  equations,  we  get 

a'  "^  h"      d'  W       b" 
the  equation  to  the  locus  of  the  point  of  contact. 

To  find  the  area  of  this  curve,  transform  its  equation  to  polar 
coordinates.     It  then  becomes 

cos^  6      sin'-^  d 

+ 


a 


b' 


r   =  c 


cos  6      sin'"^^ 


and  if  A  be  its  area,  cos''  0      sin'^  6 

'  -     -j 

A  =  2d'  (     T-^S '^cW. 

cos  6^      sm  t^N 


1848.]  INTEGRAL  CALCULUS.  185 

Let  -7  tan  0  =  tan  ^ ; 

i^  1  +  p  tan  <^ 

and  A  =  2c^  i  '  .  ,         -  sec^(bdd> 

J  ^  sec  (f)         a 

=  — T   I     (a'^  sin'''  <^  +  Z>^  cos"'  <^)  c?^ 

Again,  let  ^'  be  the  area  of  the  ellipse  which  has  the  same 
greatest  and  least  diameters,  then  if  these  diameters  be  2rj,  2?\^, 

A  =  Trr/^- 

cos''  0      sin"''  6 
4     "I"     74 
JNow   r  =  c 


a' 


=  -5T5  («'  cos'''0  +  &'''  sin'*^)  («■''  sin''*^  +  h"  cos'^0) 


cos  20 


d'b'  (V     2     /        V     2 

The  maximum   and  minimum  values  of  r  will  be  got  by 
putting  cos20  =  0  and  1  successively; 

•*■   "*>  ~  a*       2       '     ''•'  ~  "■' 

.,  _  TTC^  d'  +  h' 
*'•  ~ab    ~~2~  ' 

the  same  value  as  that  previously  got  for  A. 


18<i 


SOLUTIONS   OF   SKNATE-HOUSE    PROBLEMS. 


[1848. 


3.    Prove  that  the  remainder  after  n  tenns  of  the  infinite 
series  —  +  —  +  —  +.••  whore  a  >  1  lies  between -r^ — -tt^i  ■> 

f  =  I      —  )  and  -, w 73-^ ,  approaehiner  much  nearer  to  the 

V     j^^yx'j  (a-l)(«  +  ir"    ^^  ^ 

former  limit  when  n  is  large. 


(1).   In  general 


1  1 

+ 


^,«  +  °L(^)  ,n-Y  +. 


[m—pY      [m  +  2^) 


> 


Now 


+ 


1 


+...+ 


{n  +  l-rj))"       [n  +  1  -  {r  -  1)2)Y   {n  +  iy 

1 


+... 


+ 


+ 


+  l  +  {r-l]2)Y       {n+l+rjjy 


+ 


+  l—rpY       {n+l  +  ypY 


+ 


l{n+l-  (r-l)|>]  = 


+ 


1 


{71  +  1  + {r-  1)2)Y_ 
1 


+...+ 


{n  +  iy 


>  -, r-  +  7 rr-  +•••+  -, 7T- 1  from  above, 

[n  +  \Y       [n+lY  {n+\Y^  ' 


2/-  +  1 

^  [n+lY' 


P 


+ 


P 


"    {n+l-rpY       [n+l-[r-l)2iY 
Now  let  f'p  =  \  -  Pi  then  this  becomes 
P  ,  P  ,       ,  i^ 


+...> 


212?  +  pj 

{n  +  lY 


r«  + 


+...+ 


{n  +  i  +2)Y       [n  +  ^  +  22)Y  {n  +  i  -pT       {»  +  1)'" ' 

therefore,  a  fortiori^ 


P 


+  ...+  T 


P 


("  +  i+7?)^     ■■■      (n  +  i)*      [n  +  lY 


1848.]  INTEGRAL  CALCULUS.  187 

Let  j9  =  dx^  then  this  becomes 

r"^<&  1 


/ 


> 


n.h^'^     (/^  +  1)^' 

similarly  j^^^^->^^^^^, 

>  

We  shall  thus  obtain  an  infinite  series  of  inequalities  similar 
to  the  above.     Adding  them  all  together,  we  get 

"^   dx  1  1  1 

+  7— T^r,  +••■ 


r  dx_ 


„,_         (a-l)(«  +  ir       {n+iy       (n  +  2) 
Again,  if  ^j  —  ^iV  being  any  integer, 


+  ^"~ri  Tirvr  +•••  (/  te™is) 


(/t  +  1  +i^)"       (»  +  1  +  2^7)^ 

<  -;^ — - — c-  +  ,    ^r-  +...  (;»'  terms), 
(«+l)"      («+l)'  ^^  ^' 


1 
< 


(n  +  1)^ 

For  />  write  dx^  then  this  becomes 
^''+'''  ^  1 


/ 


whence,  as  above,  we  get 


'"  ^  _  1  1 

^  ~  (a-l)(«  +  l)'''  ^  («  +  1)^  ^  (7i  +  2) 


[      —  = 


+  7Z-n7V.+-"5 


We  have  /       — -  =  -, rr  ^7 — -Tvsrr  -  7 — rsv^l  ? 


1        r       1 


.V     2,1+2)  V^^2«+2 


2  f  a  -  1 


(a-l)(n+l)'-*    (2n4-2 

(a-l)(a-2)(a-3)         1 
■^  6  (2n  +  2)''^' 

1  («-2)(«-3)  . 


(/?+ 1)'     24(«  +  i; 


188  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

r^  dx 1__  ^  (a-2)(«-3) 

R .  r  ^  -  _L_  \ \ L_i 

•'„.  ^""(«-i)  t("+ir   (n+2rf' 

1  -T TT^U 


(a-1)  (n+ir  i        ^1 ,      1 

n+  1 


a-1  (a-l)(a-2)       1 


a-1  (w+ip  i(n  +  l)  2  (n+1 

1  a-2  1 


(w  +  1)'  2       (w+1)^"^ 

""^•■^^a;  a-2 


+. 


\ r 


rx+- 


(n  +  ir      J„,,  a-       2(,i  +  i; 
It  hence  appears  that  when  n  is  large,  -. p-  approaches 

r"+l  ^:p  r""^^  dx 

much  more  nearly  to  the  limit  I       —  than  to  I      — ,  whence 

the  latter  part  of  the  proposed  theorem  readily  follows. 

4.  If  f[x)  be  positive  and  finite  from  x  =  a  to  x  =  a  +  h^ 
shew  how  to  find  the  limit  of 

for  n  =  00  ,   and  prove   that   the   limit  in  question  is  less  than 

T  I     f{^)  ^^^)    assuming  that   the  geometric  mean  of  a  finite 

number  of  positive  quantities  which  are  not  all  equal  is  less  than 
the  arithmetic. 

Hence  prove  that  s-'^o'"'*  <  /„  £'"'*,  unless  u  be  constant  from 
£C  =  0  to  £C  =  1. 

Let  log/(ir)  =  -Fix),  then 

log  {/(«)/(«  +  ',  a)  .../(«  +  '^  hj^\ 
=  ?/  suppose. 


1849.]  INTEGRAL   CALCULUS.  189 

When  n  is  infinite,  let  -  =  dxi  then 
'        n 

7/  =  ^  {F{a)  +  F{a  +  dx)  +...+  F{a  +  h)}, 

1    /•' 
=  J  I    F{a  +  x)  dx ; 

...{/W/(.,  +  i  ;,).../(„ +  ^A)f 

approaches  to  the  limit  e^-^"  iog/t««)'''  ^^  ^^f„  iog/(«)rf«^  -^^^  smce 
the  geometric  mean  of  a  finite  number  of  positive  quantities 
which  are  not  all  equal,  is  less  than  the  arithmetic, 

{/(«)/(«  + i  a). .,/(«  +  !i^^  a)}* 

as  long  as  n  is  finite,  and  this  will  hold  up  to  the  limit  when  n 
is  indefinitely  increased :  but  in  that  case 

therefore  the  required  limit  <  y  I     f[x)dx. 

Hence  if /(o^)  =  e",  «  =  0,  and  ^  =  1,  s-^i""'  <  /^s"(7a',  unless 
?i  be  constant  from  x  =  0  to  ic=l,  in  which  case  they  are 
equal. 

1849. 

1.   Investigate  the  series 

&'      tt'  ^      cos  20      cos  30       „ 

for  values  of  0  between  —  ir  and  tt. 

Let  COS0  —  ^cos20  +  |co830  —  ...  =  n, 

sin0  -  ^sin20  +  ^cos30  -  ...=  v, 


lyO  SOLUTIONS   OF   SKNATE-IJOUSE   PROBLEMS.  [1849. 

then  if  0  lie  between  —  tt  and  tt, 


U  +-iv  =  £-*^  -  i£-*'^'  +  ^3 


*, 


=  l0g(l+£-^)  =  l0g(£     ^U  £-')  +  l0g£-*S 

=  log(2cosi0)  +  -H^; 
therefore  equating  imaginary  parts, 

V  =  sin0  -  ^sin20  +  |cos30  -...  =  ^0: 

integrating  with  respect  to  0, 

11  R^ 

-  COS0  -f  -5  COS20  -  -2  COS30  +  ...  =  -+  (7. 

To  determine  the  constant,  put  0=0; 

I  1  ^ 

Now  1  -  |.  +  ^,  -■••=  1  +  I  +  y  +••■-  2  (j!  +  p  +••• 

II  ,  /.        1        1 

.       =l+2^.  +  3.+-.-Hl  +  5.  +  3.+. 

..osi„e=«|l-(|)]{,-(l)}..., 

and  sind  =  6  —  — — ^  +... : 

equating  coefficients  of  ^^, 

1    /.        1        1 


fl'2  TT^  1  1 

and  —  =T7;  —  cos0  +  ^  cos 20  —  -r,  cos20  +•... 
4       12  2''  S'' 

2.    If  a  line  be  drawn  through   the  centre  of  an   ellipse, 
cutting  the  major  axis  at  an  angle  0,  and  the  curve  at  an  angle 


1849.]  INTEGRAL  CALCULUS.  191 

</),  (1)  prove  that 

(a^  -  h')  cos  [20  -</))  =  («■'  +  h')  cos</) ; 

and  (2)  that  f'  (ficie  =  y  . 

(1).  Let  the  coordinates  of  the  point  where  the  straight 
line  meets  the  ellipse  be  «cosa,  5  sin  a;  then  will  the  equation 
to  the  tangent  at  that  pomt  be 

cos  a  sin  a 

X  H T-  y  =  1. 

a  h     " 

Hence,  by  the  conditions  of  the  problem, 

tant/  =  -  tana, 
a  ' 

T      tan0  +  tan  6  h 

^"*^    A 1 — /]  .      .  = cota. 

1  —  tany  tan  9  a 

Hence,  eliminating  a, 

tan'd  +  tan0  tan0  _       W 
1  -  tan0  tan</)~~  ~  ~  a" 

U  -  -)i  tanfl  tan</>  =  -  f ^  +  tan'^^  ; 

,  _  _V  cos"'  B  +  a  sin'  0 
•■•    ^^"*P~  "(«•''-//■')  sine  COS© 

^(a-'+^>'-')(cos''g+sin-'0)-(ff''-^/)(cos'''e-siu-'g) 
2(«'''-Z>-)  sin  0  cos  0 

_  fl--' +  ^-^  -  (^-^  _  ^^)  C03  2g 
(«■■'- 6^)  sin  20  ' 

.-.    (a'  -  6')  (sin20  tan<^  +  cos20)  =  d'  +  //, 

and    («'  -  />')  cos  {29  -  (f))  =  («'  +  Z»"')  cos<^. 

(2).    Again,  since 

_      b""  cos' 0  +  a"  iiW  9 
^"^9--  {d'-h-^)^[n9cos9 ' 


192  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

=  /     X?n7r  -  tan     -r-r^ — ,2.    .    ^ ^h  at/ (2) 

1   C"' 
=  -  I     onvdO  by  adding  (1)  and  (2) 

where  ?«  is  a  constant  integer  to  be  determined. 

For  this  purpose  we  observe,  first,  that  its  vahie  is  inde- 
pendent of  any  relation  between  a  and  h  •  and  secondly,  that  if 
a  =  b  the  ellipse  becomes  a  circle  and  (jy  always  =  ^tt.  Hence 
in  this  case 

(Pde=        ^dB  =  —  and  m  =  -  1. 
Hence  also,  in  all  cases,  m  =  —  1  and 


f 

J  0 


^dB  =  -- 


3.  Through  a  given  point  B  (fig.  84)  of  the  axis  of  a;  a  line 
is  drawn  parallel  to  the  axis  of  y :  to  any  point  Q  of  this  line 
another  straight  line  is  drawn  from  the  origin  and  produced  to 
P  until  PQ  =  BQ.  Find  the  equation  to  the  locus  of  P,  trace 
the  cui've,  and  find  the  whole  area  included  between  the  cm've 
and  the  asymptote. 

Extend  the  geometrical  description  so  as  to  include  the  whole 
of  the  curve  given  by  the  equation. 

Let   AB  =  a,     AP  =  r,     PAB  =  0.     Then 

QP  =  r  —  a  sec  B, 

QB  =  a  tan  0  ; 

.*.    r  =  a(sec0  +  tan0), 

the  polar  equation  to  the  curve. 


1849.]  INTEGRAL  CALCULUS.  193 

DifFcrentiating, 

-jji  =  asccO  (t&nO  +  secO) 
da 

=  r  sec  0, 

.".    r^ -J- =  rco&O  =  a  (1  + sinO). 
dr 

When  0  =  -  and  —  ,  r  =  oo  and  r^  -^  =  2a  and  0 :  hence 
2  2'  dr  ' 

the  axis  of  ?/,  and  the  line  DCD  parallel  to  it  such  that  AC  =2a^ 

are  asymptotes  to  the  curve : 

f\      r.  I   dr       ^ 

^  =  ^'  '•  =  ^'   rTe  =  ^^ 

0  >  0,     <  2  "■?     ^  i^  positive  ; 

„      TT  1  +  sin  0  . 

t/  >  —  <  TT,     r  =  —  a J.  —  IS  negative ; 

2  '  COS0  ^  ' 

rt  Stt  1  -  sin  0 . 

t/>7r<— -,     r  =  —  a jr—  is  nesrativc  : 

2  '  COS0  ^  ' 

B  =  ^Stt,      r  =  cc  , 

„       Stt       ^  1  -  sin0  . 

0  >  —  <  27r,     r  =  a ^  -  is  positive  : 

2  '       •  cost^  ^ 

the  negative  values  of  0  give  no  new  branch  of  the  curve. 
Hence  the  cun^e  is  of  the  form  represented  in  (fig.  85). 
To  find  the  area  [A)  included  between  the  curve  and  the 
as}Tnptote  D  CD'.     Produce  AF  to  meet  the  asymptote  in  R ; 
then  the  element  of  the  area  A 

BA  =  ^{AE'  -  AF')  8A  =  ^pasecOy  -  a'  (sec0  +  tan0f}  SO 
=  ^a"  (3  sec' 0  -  2  sec 0  tan  0  -  tan" 6)  BO 
=  ^d'  (2  sec'd -  2  sec0  tan0  +  1)  SB  ; 

.-.  A  =  ^d'  f2tan0-  -\  +  b]  +  C 

'     \  cosy      / 


^       [  \l  +  smBj  j 


194  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

from  6  =  0  to  0  =  ^TT  gives  ^A, 

.-.  ^  =  (^7r-f  2)  d\ 

the  required  area. 

If  P'  be  the  point  where  AP  cuts  the  branch  BP'  of  the 
curve,  it  is  evident  from  the  tracing  of  the  curve  that  QP  =  QB : 
hence  the  curve  may  be  described  as  the  locus  of  the  point  P 
on  the  Hne  AR  whose  distance  from  Q  equals  QB. 


(     195     ) 


GEOMETRY  OF  THREE  DIMENSIONS. 

1848. 

1.  If  three  chords  be  drawn  mutually  at  right  angles 
through   a  fixed  pomt  within   a  surface   of  the    second    order 

whose  equation  is  u  =  0,  shew  that  2  ^-  will  be  constant,  where 

R  and  r  are  the  two  portions  into  which  any  one  of  the  chords 
drawn  through  the  fixed  point  is  divided  by  that  point. 

Prove  also  that  the  same  will  be  true,  if  instead  of  the  fixed 
point  there  be  substituted  any  point  in  the  sm'face  whose  equa- 
tion is  w  =  c. 

We  shall  prove  the  second  part  of  this  only,  since  it  mani- 
festly includes  the  first. 

Let  the  equation  to  the  surface  a  =  0,  referred  to  its  centre 
and  axes,  be 

Ax"  +  Bf  +  C£'  =  1 (1). 

Let  a,  /8,  7  be  the  point  through  which  the  lines  are  drawn ; 
then,  since  it  always  lies  on  the  surface  u  =  c,  we  have 

Ad'  +  ^/3'^  +  Ct^  =  1  +  c (2). 

Let  l^m^n^^  ^i^h^t  h^^^a'hi  ^^  ^^^  direction-cosines  of  the  lines, 
then  their  equations  are 

-,— =  ^ -=  -  =  Px  say 3), 

h      ~     -\     ~      >\     ~^' ^^' 

xj-a      y  -  ^      z-y 

^'^  =^r=''' <"' 

02 


106  SOLUTIONS  OF   SENATE-HorSI-:   PROBLEMS.  [1848. 

('.»'.»,)7  (^.;'».2*'.2)^  ih"',",)  lacing,  since  (3),  (4),  (5)  are  at  right 
angles  to  one  another,  subject  to  the  conditions 

K'  +  f:  +  K  =1 (6), 

<  +  <  +  <=! (7), 

<  +  <  +  <  =  1 (8). 

Where  (3)  meets  (1)  we  have,  substituting  for  xyz  in  tenns 

^  (a  +  /.p,)'^  +  ^  (/3  +  m,p^Y  +  ^  (7  +  n.p.Y  =  L 
The  roots  of  this,  considered  as  an  equation  in  p^,  are  R^r ;  hence 

Rr  ~  Aa'  +  ^y8^  +  Cy'  -  1 

^  Ai;'  +  ^m;-^  +  c^;^ 

c 
Similar  expressions  resulting  from  (4)  and   (5),  we  get  by 
(6),  (7),  (8), 


^ 1       A+B+C 
^  Rr~  c  ' 


which  is  constant. 


2.  Find  the  locus  of  the  foot  of  the  pei-pendicular  let  fall 
from  the  origin  on  the  tangent  plane  to  the  surface  xyz  =  a^ ; 
point  out  the  general  form  of  the  required  sm'face,  and  find  the 
whole  included  volmue. 

The  equation  to  the  tangent  plane  to  the  given  surface  at 

a  point  {xyz),  is 

^,      V,      ^, 

^  +  ^  +  -  =  3. 

xyz 

The  equations  to  the  pei'pendicular  on  this  plane  from  the 
origin,  are 

At  the  intersection  of  these  we  have 

^^1  =  yy^  =  ^^1  =  - — 3^ ; 

therefore,  since  xyz  =  a^,  we  get  as  the  equation  to  the  locus 
required,  (^^-^  +  y^^  +  ^-^  =  21a'x^y^z^. 


1848.]  GEOMETUY    OF    THREE    DIMENSIONS.  197 

The  form  of  this  surface  will  be  that  of  four  similar  sym- 
metrical pear-shaped  portions,  meeting  in  a  point  at  the  origin, 
and  lying  in  the  octants  +  +  +,  -\ ,  -  +  — , h- 

The  equation  to  the  surface,  transfonned  to  polar  coordinates, 

becomes 

r"  =  27rt^  cosO  slu^0  cos^  sin^. 

And  if  V  be  the  volume  of  one  of  the  portions 
V  =  ^JfJr'  smSdrdddi/, 

between  proper  limits, 

9    .    /-*"/■-" 
=  -  aM     I      COS0  sin'0  cos^  sm<f)d6d(f> 

2  J  0    -'  0 


9        ; 
=  -  a^  j     cos  (p  sin  (p  dcj) 


=  ^  a' 
16 


therefore  If  V  be  the  whole  volume  of  the  surface, 

V=  4F'  =  -a^ 
4 

3.  A  plane  moves  so  as  always  to  enclose  between  Itself 
and  a  given  surface  S  a  constant  volume ;  prove  that  the 
envelope  of  the  system  of  such  planes  is  the  same  as  the  locus 
of  the  centres  of  gravity  of  the  portions  of  the  planes  comprised 
within  S. 

Conceive  the  plane  to  receive  a  small  twist  about  any  straight 
line  passing  through  the  centre  of  gravity  of  the  portu)n  com- 
prised within  S;  then,  whatever  portion  is  cut  off  from  the 
enclosed  volume  on  one  side  of  this  line,  an  equal  portion  will 
be  added  to  it  on  the  other,*  so  that,  by  the  conditions  of  the 
problem,  the  plane  will  pass  from  any  one  position  to  the  con- 
secutive one  by  turning  about  a  line  passing  through  the  centre 
of  gravity  of  the  portion  comprised  within  S.  Therefore  the  en- 
velope of  the  planes  will  be  the  same  as  the  locus  of  the  centres 
of  gravity  of  the  portions  of  the  planes  comprised  within  S. 

*   See  Cambridge  and  Dublin  Mathematical  Journal,  vol.  iii.  p.  181. 


198  SOLUTIONS   OF   SENATK-IIOUSE    PROBLEMS.  [1848- 

4.  OA^  OB^  OC,  are  three  straight  Ihies  mutually  at  right 
angles,  and  a  lumuious  point  is  placed  at  C;  shew  that  when 
the  quantity  of  light  received  upon  the  triangle  A  OB  is  con- 
stant, the  cui"ve  which  is  always  touched  by  AB  will  be  an 
hyperbola  whose  equation  referred  to  the  axes  OA^  OB^  is 
[y—mx)  (x  —  my)  =  m&^  where  OC  =  c^  and  m  is  a  constant 
quantity. 

With  C  as  centre,  and  CO  as  radius,  describe  a  spherical 
surface,  then  the  quantity  of  light  received  on  the  triangle 
A  OB  Is  the  same  as  that  received  by  the  spherical  triangle 
CA'B'  intercepted  between  the  planes  COA^  COB^  CBA^  and 
will  therefore  be  proportional  to  the  area  of  that  surface.  But 
if  S  be  this  area, 

S  =  ^irr'  {A  OB'  +  OAB'  +  OB' A  -  it) 
=  27rr'  ( OAB'  -f  OB' A  -  ^tt), 

since  A  OB  is  a  right  angle. 

Therefore  if  the  quantity  of  light  received  by  the  triangle 
be  constant,  OAB'  +  OB' A  must  be  so,  =  2a  suppose. 

Let  the  angle  OB' A  =  a  +  0^  then  OAB'  will  =  a-d^  and 
the  equation  to  the  plane  ABC  will  be 

cos(a+^)  x-\-  cos[a  —  6)y+  {1 -cos'(a  + ^) -cos''(a- ^)}*^=7>; 

p  will  be  detennined  from  the  consideration  that  where  this 
meets  the  axis  of  s,  we  have  s  =  c ; 

.-.  {l-co3'''(a+^)-cos'(a-^)li  =  c, 
therefore  the  equation  to  AB  is 

cos {c(.+  6)  X  +  cos[a—  6)  y  =  (1  —  cos'"' {a+  6]  —  cos^ (a -  6)]^  c 

=  (-cos2acos2^)4  c (1). 

The  quantity  (— cos2acos^)*  is  real,  since  2a  is  greater  than 
a  right  angle  and  less  than  two  right  angles,  and  therefore 
cos  2  a  negative. 

Putting  tana  =  w,  tan^  =  t,  (1)  becomes 

(1  -  7it)  x+  [l  +  nt]  y  =  {{n'  -  1)  (1  -  f]}^  c, 

and  we  have  to  find  the  locus  of  ultimate  intersections  of  this 
line,  subject  to  the  variation  of  f. 


1848.]  GEOMETRY   OF   THREE   DIMENSIONS.  199 

Clearing  the  equation  of  radicals  and  arranging  according 
to  powers  of  f,  it  becomes 

f  [re  [x-yf  +  in'  - 1)  c'}  +  2tn  [if-x')  +  (x  +  yf  -  {n'  - 1) c'  =  0. 
Eliminating  t  between  this  equation  and  its  derivative,  we  get 

{[x  +  yY  +  {n^  -  1)  c']  [n'  [x  -  yY  +  {n^  -  1 )  c'^}  =  n^  (f  -  xj, 
which  may  be  reduced  to 

[x  +  yr-re[x-yf={n^-\)c\ 

or  {(1  -n)x  +  (1+  n)  y}  {(1  +  n)  ic  -I-  (1  -  n)  y\  =  («'  -  1)  c' ; 

1        f              .       n  —  \ 
therefore  puttmg =  »i, 

[y  —  mx)  [x  —  my)  =  hik? 

is  the  equation  to  the  curve  always  touched  by  AB. 

In  a  manner  similar  to  this  may  be  solved  the  following 
problem,  set  in  1851. 

Let  a  spherical  surface  whose  centre  is  the  origin  of  coor- 
dinates meet  two  of  the  coordinate  planes  in  the  great  circles 
Zx^  Zy\  also  let  the  points  P,  Q  be  taken  m  Zx^  Zy  respec- 
tively, so  as  to  make  the  surface  of  the  spherical  triangle  PZQ 
constant:  shew  that  the  curve  which  is  always  touched  by  the 
great  circle  PQ  has  for  its  equations  x^  -\-  y^  -{■  ^  z=  d^^  and 
xy  =  ^d^  sin^,  where  E  is  the  spherical  excess  of  the  triangle 
PZQ. 

The  geometrical  conditions  of  this  problem  are  the  same  as 
those  of  the  foregoing.  Writing  z  for  c,  we  have  as  the  equation 
to  the  surface  always  touched  by  the  plane  through  the  centre, 

2  7r  +  1  ^        ^         ' 


+ 
x'  +  /  +  z' 


cos2a. 


200  SOLUTIONS    or    Si:NATl->Ht>i:SE    PROBLEMS.  [1848. 

But  -rr  +  E=  A' OB'  +  OAB'  +  OB' A  In  the  previous  notation 
=  -^TT  +  2a ; 
.-.  2a  =  ^TT  +  E, 
and  our  equation  becomes 

xy  =  \  (a;"'*  +  ^^  +  2:''')  sinE". 

But  since  the  curve  is  traced  on  the  sphere,  w^e  have 
x^  -\-  y^  +  z^  =  a^,  and  we  get  as  the  equations  to  the  curve, 

X  +  y   '\-  z  =^  a  , 

xy  =  ^a"  sinjE", 
the  required  equations. 

5.  If  0  be  a  given  point  in  a  surface  of  the  second  order, 
and  OA^  OB,  00,  any  three  chords  passing  through  0  mutually 
at  right  angles,  shew  that  the  plane  ABC  will  always  pass 
through  a  fixed  point. 

Take  0  as  origin,  and  the  three  lines  OA,  OB,  OC,  xn.  any 
position  as  axes ;  let  the  equation  to  the  surface  be 

Ax'->rBy'+Cz'-\-2Ayz+2B'zx-\-2C'xy+'iA"x+W'y+2C"z={). 

Then  the  length  of  OA  will  be  the  value  of  x,  when  y  =  0, 

2  =  0;  hence 

9  A" 
0A  =  -^-: 
A 

2B"                     2C" 
similarly    OB  = ^- ,     OC  = y^ , 

and  the  equation  to  ABC  will  be 

Ax      By       Cz      ^      ^  ,,, 

y+^  +  ^,  +  2  =  0  (1). 


And  the  equations  to  the  normal  at  0  are 
_    3/     _ 


2 A"  ~  2B"  ~  2C"~  2 [A""  +  B"^  +  C"''f 

where  r  is  the  distance  from  the  origin  of  the  point  [xyz). 


(2), 


1849.]  GEOMETRY    UF   THREE    DIMENSIONS.  201 

Where  (I)  and  (2)  intersect,  we  have,  dividing  eaeli  term 
of  (1)  by  the  corresponding  member  of  (2), 

r 

Now  we  may  establish  one  relation  among  the  nine  coeffi- 
cients of  the  equation  to  the  surface.  Let  then  [A"^  +  B'"^  +  C"'^) 
be  constant,  then  the  above  equation  shews  that  r  varies  inversely 

as  ^  +  5  +  a 

But  it  is  known  that  if  the  equation  be  transformed  into  the 
form 

Px'  +  Qy'  +  Bz'  +  2P"x  +  2Q"y  +  2B"z  =  0, 

the  quantities  P,  Q^  B,  are  the  roots  of  the  equation 

(-S'-  A)  [S-  B)  {8-  C)  -  A"  [S-A)-  B"  [S-B]-  C"  {S-  C) 

-  2A'B'C'  =  0. 

Hence,  by  the  theory  of  equations, 

A+B+C=P+Q  +  B,  a  constant. 

Hence  ?*,  the  distance  from  0  of  the  point  in  which  the  plane 
ABC  intersects -the  normal  at  0,  is  constant,  therefore  the  plane 
ABC  always  passes  through  a  fixed  point. 

1849. 

1.  If  planes  be  drawn  through  any  two  generating  lines 
of  an  hyperboloid  which  intersect,  shew  that  they  will  cut  the 
surface  in  another  pair  of  generating  lines. 

Let  the  equation  to  the  hyperboloid  be 

'2  2  2 

-+^--=1  (1) 

Now  a  plane,  drawn  through  two  intersecting  generating 
lines  of  an  hyperboloid,  touches  the  hyperboloid  at  their  point 
of  intersection.  Let  then  x\  y\  z\  be  the  coordinates  of  this 
point,  then  the  equation  to  the  plane  will  be 


XX      yy      zz 

0 


+"/-^:  =  i c^). 


202  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 


a;',  y\  z\  being  subject  to  the  condition 
or       h'       c 


^+^-:;ii  =  i (3), 


(1)  +  2  (2)  +  (3)  gives 

m-m'-m- (^)- 

a  condition  which  must  be  satisfied  by  the  coordinates  of  any 
point  where  (2)  intersects  (1). 

Now  (4)  may  be  put  into  the  form 

ic  +  aj'Y'*     .  _  (z-\-z'^     (y-^y 


a 
which  may  be  written 

shewing  that  where  (2)  meets  (1)  we  have  either 
fi±£:+2=i.fyif:  +  y±l]  and  ^^-2  =  \  {'-±i  -  ^f]  , 

O/  \C  U       J  Ct  rC  \      C  0      J 

representing  one  generating  line,  or 

^E±ii  +  2  =  i' (-£±1  _  2^)  and  i^^' -  2  = '.,  f  ^  +  ?t±l^) 
a  \    c  0    J  a  fc   \    c  b    ) 

representing  another. 

Hence  if  planes  be  drawn  through  any  two  generating  lines 
of  an  hyperboloid  which  intersect,  they  will  cut  the  surface  in 
another  pair  of  generating  lines. 

2.  If  u  =f[x^  y,  z)  be  a  rational  function  of  cc,  y,  0,  and 
if  w  =  0  be  the  equation  to  a  surface,  for  a  point  («,  J,  c)  of 
which  all  the  partial  diiferential  coefficients  of  u  as  far  as  those 
of  the  (w  —  1)*  order  vanish,  shew  that  the  conical  sm-face  whose 
equation  is 

[(x-o)  ^  +  (3/-^)  I  +  (^-^)  j}  /(«,  h  c)  =  0, 
will  touch  the  proposed  surface  at  the  point  (a,  J,  c). 


1849.]  GEOMETUY    OF   THREE    DIMENSIONS.  203 

Tx                          X  —  ay  —  hz  —  c  ,. 

Let  -^j—=^ =  — — 1 

/  111  /rt  ^     ' 


be  the  equations  to  any  line  passing  through  (a,  Z>,  c). 

Denoting  each  member  of  ( I )  by  r,  we  shall  obtain  the  other 
points  of  intersection  of  (1)  with  u  =  0  by  writing 

a  -f-  Ir  for  cc,     h  -\-  mr  for  y,     c  +  nr  for  z 

in  the  equation  u  =  0.  This  gives,  developing  by  Taylor's 
Theorem, 

7  HI  7  HI  7  n« 

which,  since  -^  =  -^  =  -j-y^^  =  0  for  all  values  of  m  less  than  n 

becomes,  dividinsr  out  by , 

'  ^  ^  1.2. ..w' 

If  the  line  (1)  touch  the  surface  ?<  =  0  at  the  point  («,  J,  c) 
equation  (2)  must  be  satisfied  by  making  r  indefinitely  small; 
(2)  will  then  become 

a  condition  to  be  satisfied  by  the  direction-cosines  of  (1)  in 
order  that  it  may  touch  m  =  0  at  the  point  (a,  5,  c).  To  obtain 
the  locus  of  all  such  lines,  we  must  eliminate  /,  w,  n  from  the 
above  equation  by  means  of  (1).     This  gives 

j(^-a)  I-  +  (2,-i)  I  +  (.-  c)  gV{a,  J,  c)  =  0 

as  the  equation  to  the  conical  surface  which  touches  u  =  0  at 
the  point   (^/,  />,  c). 


204  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

3.  A  rod  AB  is  fixed  to  a  universal  joint  to  A^  and  another 
rod  BP  is  conneeted  to  it  by  a  universal  joint  at  B:  all  di- 
rections of  the  rod  being  equally  probable,  find  the  chance  of 
P  lying  between  two  spherical  surfaces  of  given  radii,  whose 
connuon  centre  is  A ;  and  shew  that  the  chance  of  P  lying 
within  a  given  elementary  portion  of  space  containing  the  point 
P,  varies  inversely  as  AP^. 

Let  AB  =a,BP=  h. 

The  chance  that  P  will  lie  between  two  spherical  surfaces 
of  given  radii,  is  the  chance  that  the  angle  ABP  will  lie  be- 
tween two  values  6^  and  6^,  which  correspond  to  the  values 
r^  and  r^  of  -4P,  r^  and  r^  being  the  radii  of  the  spherical  shell. 

Now  the  chance  that  ABP  will  lie  between  6  and  d  +  B0 

=  area  of  zone  described  by  P  about  B  fixed,  while  0  has  all 
values  from  ^  to  0  +  B6  ^  surface  of  sphere  generated 
by  P  about  B  fixed, 

27rb  sin^  x  b  s[n0       ,    .    /,  ,/, 

4:7rb'  ^ 

Hence  the  chance  required 
1   f ^   .    ^       1,       z,  ■/,^        Ifa'  +  b'-r;'       a'  +  b'-r 

__    I        oivi  /-/  — I  ona  H     nr\a  H    \    -—  i — i^ 


-  1    sin  ^  =  -  (cos  0,  —  cos'^, 


2  /  „  2  ^         '  '  '''       2\        2ab  2ab 


4ab      ' 
The  chance  that  P  will  lie  in  an  element  V  of  space  about  P^ 
=  chance  of  falling  in  a  spherical  shell  about  A  as  centre  of 

^,  .  1  -,         ,.  volume  of  element 

thickness  or,  radius  r  x  ^ ;^-^j — n- 

'  volume  ot  shell 

_  2rSr  V      _     V       1       1 

4rtZ'       47rr''*^r       STvab '  r       r' 

4.  Determine  the  condition  to  which  the  vertices  of  a  system 
of  cones  which  envelope  an  ellipsoid  must  be  subject,  in  order 
that  the  centres  of  the  ellipses  of  contact  may  be  equidistant 
from  the  centre  of  the  ellipsoid. 


1840.]  GEOMETRY   OF   THREE    DIMENSIONS.  205 

Let  ^,  7;,  ^,  be  the  coordinates  of  the  vertex  of  any  one  of 
the  cones ;  then  if  tlie  ellipse  be  refeiTcd  to  its  centre  and 
axes,  the  equation  to  the  plane  of  contact  will  be 

1:^  +  ^  +  ^  =  1 (1). 

The  centre  of  the  ellipse  of  contact  will  be  the  intersection 
of  (1)  with  the  straight  line  joining  the  centre  of  the  ellipse 
with  the  vertex  of  the  cone ;   its  equations  are 

X       ^       z 

Hence  if  h,  A-,  I,  be  the  coordinates  of  the  centre  of  the 
ellipse, 

In  order  that  the  centres  of  the  ellipses  may  be  equidistant 
from  the  centre  of  the  ellipsoid,  we  must  have 

Ji^  +  I?  +  T^  =  constant,  p^  suppose ; 


the  equation  to  the  locus  of  the  vertices. 

5.  Determine  the  form  of  the  termination  of  a  honeycomb 
cell  on  this  principle,  that  if  a  sphere  which  will  just  pass 
through  the  hexagonal  transverse  section  be  dropped  into  the 
cell,  the  unoccupied  space  at  the  extremity  of  the  cell  shall 
be  the  least  possible.* 

Let  ahe  (fig.  86)  represent  half  of  one  of  the  rhomboidal 
plates,  three  of  which  close  each  hexagonal  cell.  ABC  re- 
present the  eighth  part  of  a  sphere.  Then,  by  the  general 
principle  of  Envelopes  (see  Cambridge  and  Dublin  Mathematical 
Journal^  vol.  iii.  p.  181),  the  volume  in  question  is  least  when 
the  point  of  contact  d  is  the  centre  of  the  rhomboid:  or  we 
must  have  ac  =  lad. 

Let  OA  =  a,  07)  =  r ;    .'.  a  =  r  cos  .30°  =  r  -  . 


*  For  this  solution  we  are  indebted  to  Mr.  Goodwin, 


206  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

Let  Oac  =  0  J    .-.  a  =  Od  =  ad  tan0  =  ^ac  tan^ 
=  ^r  cosec^  tan0  ; 
.       3*       ,        1 


COS(f>  ' 

.'.  COS  9  =  ,-T  . 

3* 
Again,  hd  =  ED  sin  60  =  r  —  ,  and  ad  =  ^r  cosecc^  ; 

.'.  t&nbad  =  ^  =  3*  sm(f>  =  3*  (1  -  i)4  =  2*. 

These  are  the  angles  required. 

6.  The  tangent  plane  to  a  surface  S  cuts  an  ellipsoid,  and 
the  locus  of  the  vertex  of  the  cone  which  touches  the  ellipsoid  in 
the  curve  of  intersection  is  another  surface  S'.  Prove  that  S  and 
S'  are  reciprocal,  that  is,  that  8  may  be  generated  from  8' 
in  the  same  mamier  as  ;S^'  has  been  generated  from  8. 

Take  the  axes  of  the  ellipsoid  as  coordinate  axes,  and  let 
the  equation  to  8  be 

^  =  0.  °  • 

That  to  its  tangent  plane  at  any  point  xt/z,  is 

,  V  d8     I  ,  68      ,         ,  d8 

This  may  be  put  under  the  form 

d8         d8         d8 

dS  dS         d8~ 

dx       "^   dy  dz 

If  ^,  77,  ^,  be  the  coordmates  of  the  vertex  of  the  cone  touch- 
ing the  ellipsoid  in  the  cm-ve  of  intersection  with  this  plane, 
we  have  ^ 

^  dx 

a'^"^         dS         dS' 
dx      ^  dy  dz 

with  similar  expressions  for  t]  and  ^. 


1849.]  GEOMETRY   OF   THREE   DIMENSIONS.  207 

Again,  if  with  [xyz)  a  point  of  S  as  vertex  we  describe  a 
cone  touching  the  ellipsoid,  the  equation  to  the  plane  of  contact 
will  be 

«l  ,  ^  ,  ^  _  1 
d'  ^  h'  "^  6'  ~    ' 

1^,  77,  f  being  its  current  coordinates.  To  find  the  locus  of 
ultimate  intersection  of  these  planes,  eliminate  .r,  ?/,  z  between 
the  differential  of  the  preceding  equation,  and  of 

this  gives  ^^  +  ^  =  0, 
h       dy 

Multiplying  these  equations  in  order  by  a?,  y,  2,  and  adding, 
^'eget  js         dS        dS      ^ 

^  +  ^;^  +  ^^  +  '^='' 
dS 
^  dx 

dx       "  dy  dz 

with  similar  expressions  for  77,  ^. 

Hence  the  locus  of  f ,  97,  ^  is  5". 

That  is,  the  locus  of  the  vertex  of  the  cone  touching  the 
ellipsoid  in  its  curve  of  intersection  with  a  tangent  plane  to  8 
is  the  same  as  the  envelope  of  the  plane  of  contact  when  a  cone 
is  drawn  from  a  point  of  8  as  vertex,  circumscribing  the 
ellipsoid.     This  holds  for  all  surfaces,  therefore  for  8' . 

But  from  the  mode  of  generation  of  8\  it  is  easy  to  see  that 
the  envelope  of  the  planes  of  contact  of  cones  drawn  from  its 
points  as  vertices  is  8\  therefore,  by  what  has  been  proved,  the 
locus  of  the  vertices  of  the  cones  touching  the  ellipsoid  in  its 
curves  of  intersection  with  the  tangent  planes  to  8'  is  8^  that 


208  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

is,  S  may  be  generated  from  S'  in  the  same  manner  as  S'  was 
from  S,  or  S  and  S'  are  reciprocal. 

1850. 

1.  If  from  a  point  0  be  drawn  any  two  lines  to  the  polar 
plane  of  0  in  a  surface  of  the  second  order  and  meet  the  plane 
in  A  and  i?,  and  if  the  central  conjugate  plane  of  OA  meet  OB 
in  C,  and  the  central  conjugate  plane  of  OB  meet  OA  in  Z), 
CB  is  parallel  to  AB. 

Take  that  diameter  of  the  ellipsoid  which  passes  through  0, 
and  two  diameters  conjugate  to  it,  as  axes.  Let  the  equation 
to  the  ellipsoid  be 

H  2  2 

X        y       z 

1-  —  -I =  1 

a         b         c 

Let  ^  be  the  distance  of  0  from  the  centre,  then  the  equation 
to  its  polar  plane  is 

a 

"  =  T- 

Let  the  coordinates  of  ^  be  — ,  ?/j  s^ ;  of  ^,  -^ ,  y^,  z^.    Then 

the  equations  to  AB  are 

a'"        y  —  V,       z  —  z^  ,. 

f  3/l   -  3^2  ^1    -  ^2 

The  equations  to  OA  will  be 

a" 
X ^ 

t       y      z 
:75 ^  =  f"  =  ^  =  ^  suppose (2), 

I  . 

therefore  the  equation  to  its  central  conjugate  plane  is 
'a"^       J\   X       w,?/       z^z 

l-f)5='  +  ¥  +  iP  =  ° <-'>■ 

Similarly,  the  equations  to  OB  are 

'2 

a 

^  —  -y 

^^  =  ^  =  i  =  .-, (4), 


1850.]  GEOMETRY   OF   THREE   DIMENSIONS.  209 

therefore  that  to  its  central  conjugate  plane  is 

At  the  point  C,  the  intersection  of  (3)  with  (4),  we  have 

1-^ 


Similarly,  it  will  be  seen  that  at  i),  the  intersection  of  (2)  and 
(5),  we  have 

a 


'2  /      '•!  , 

the  values  of  .r  at  C  and  D  are  each  equal  ^o  -g  +  (^  -  I)  »',? 
therefore  the  equations  to  CD  are 


«^  =  T  +  h--^ri' 


I     VI       J  ''    y,-y^      ^1  -  ^2  ' 

by  comparing  which  equations  with  (1)  we  see  that  CD  is  parallel 
to^i?. 

2.  A  plane  moves  so  as  always  to  cut  off  from  an  ellipsoid 
the  same  volume ;  shew  that  it  will  in  every  position  touch  a 
similar  and  concentric  ellipsoid. 

If  a  plane  be  drawn  touching  the  interior  of  two  similar 
and  concentric  ellipsoids,  the  point  of  contact  will  be  the  centre 
of  its  elliptic  section  made  by  the  exterior  one.  Now  conceive 
this  plane  to  receive  a  small  twist  about  any  diameter  :  it  will 
still  remain  in  contact  with  the  interior  ellipsoid,  and  whatever 
portion  is  taken  from  the  volmne  intercepted  between  it  and 
the  exterior  ellipsoid  on  one  side,  will  be  added  to  it  on  the 
other,  therefore  that  volume  will  be  unaltered.  Hence  con- 
versely, it  follows  that  if  a  plane  move  so  as  always  to  cut  off 

p 


210  SOLUTIONS   OF   SENATF.-IIOUSE   PROBLEMS.  [I8o0. 

from  an  ellipsoid  the  same  volume,  the  sm*tacc  which  it  always 
touches  will  be  a  similar  and  concentric  ellipsoid. 

3.  If  F{x^  y^  c)  =  0  be  the  equation  of  a  system  of  curves, 
where  c  is  a  variable  parameter,  and  (f)  (.r,  y)  =  0  the  equation 
of  the  envelope  of  the  system ;  shew  that  ^  (.x,  y)  =  0  is  the 
equation  of  a  cylmder  whose  intersection  with  the  surface 
jP(.r,  y,  s)  =  0  is  the  locus  of  points  which  in  sections  parallel 
to  the  planes  of  yx^  zx^  have  their  tangents  parallel  to  the 
axis  of  z. 

Ex.  The  cone  whose  equation  is  a?  +  y'^  +  s^  =  [Ix  +  my  +  nzj 
is  cut  by  planes  parallel  to  the  planes  of  yz  and  zx ;  find  the  loci 
of  the  extremities  of  the  diameters  of  the  sections  which  are  con- 
jugate to  the  vertical  diameter. 

(a).   The  equation         (f>{x,y)=0 (1) 

results  from  the  elimination  of  c  between  the  equations 

F{x,y,c)  =  0, 

and   -^-  =  0. 
dc 

It  will  therefore  be  also  obtained  by  eliminating  s  between 

F{x,y,z)=^0 (2), 

and    -^  =  0. 
dz 

Hence,  where  the  sm-faces  represented  by  (1)  and  (2)  inter- 
sect, we  have 

f- (3). 

Now  the  equation  to  a  tangent  plane  to  (1),  parallel  to  the 

plane  of  yz.  is 

,         ,dF     ,         ,dF     ^ 


1850.]  GEOMETRY   OF   THREE    DIMENSIONS.  211 

dF 

If  tlicrcforc  -^-  =  o,  this  becomes 

az 

which  is  evidently  parallel  to  the  axis  of  z. 

Hence  at  the  curve  of  intersection  of  the  cylinder  (1;  and 
the  surface  (2),  the  tangent  in  a  section  parallel  to  the  jdane 
of  yz  is  parallel  to  the  axis  of  z. 

Similarly  it  may  be  shewn  that  the  tangent  in  a  section 
parallel  to  the  plane  of  xz  is  parallel  to  the  axis  of  z. 

(/3).  In  the  example,  the  tangent  at  the  required  points  are 
parallel  to  the  vertical  diameter,  tliat  is  to  the  axis  of  z^  hence 
we  get  the  locus  required  by  eliminating  z  between 

F{x^  y,  z)  =  oc'  +  /  4-  z'  -  [Ix  +  my  +  nzf  =  0, 

dF 

and  -T-  =  2z  —  2n  [Ix  +  my  +  nz)  =  0. 

The  latter  equation  gives 

Ix  +  my 
1  -  n^ 

Hence     a.'"  +f=  [Jx  +  my;  (l  +  ,^^  -  n'  ^^^^1' 

_  {Ix  +  myY 

is  the  equation  to  a  cylinder,  whose  intersection  with  the  given 
surface  is  the  required  locus. 

4.  If  .r,  ?/,  2-,  be  the  coordinates  of  any  point  P  on  the  surface 
f{x,  y,  z)  =  0,  x\  y\  z  of  a  point  F  on  the  surface /(a?',  y\  z')  =  0, 
and  for  any  position  of  P,  x'  =  Ix^  y  =  my^  z  =  nz ;  and  if  the 
surfaces  be  such  that  when  we  take  any  two  points  P,  Q  on  the 
first  and  two  corresponding  points  P',  Q'  on  the  second,  PQ  is 
equal  to  P  ^ ;  find  the  fonn  of  the  sm^aces. 

Let  f ,  17,  ^  be  the  coordinates  of  Q. 
Then  ?f ,  W17,  nf  are  those  of  Q\ 

...  PQ-  ={x-  I^Y  +{y-  mriY  +  [z  -  nXf, 
PC/  =  (I  -  Ixf  +  (7;  -  my?  +  (^-.  nz)% 

P2 


212  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

and  these  are  equal ;  hence  we  get 

(^  -  I^y  +  {y-  mvT  +{z-  n^Y  =  (^  -  Ixf  +  (77  -  m^/f  +  {^-  nz)\ 

oi'{l-r')x'+{l-ni')f+{\-7{')z'={\-r')^'+{l-m')v'  +  {l-n')^\ 

for  all  values  of  x,  3/,  z ;  |,  ■»;,  ^,  consistent  with  the  equation  to 
the  surface /(.r,  y,  z)  =  0. 
We  must  therefore  have 

(1  -  r)  x'  +  (1  -  m')  y'  +  (1  - '?'")  z'  =  constant,  a'  suppose, 
which  determines  the  form  of  the  required  surfaces,  which  are 
evidently  central  smfaces  of  the  second  order,  of  which  the  axes 
of  coordinates  are  principal  axes. 

5.  It  is  not  possible  to  fill  any  given  space  with  a  number 
of  regular  polyhedrons  of  the  same  kind  except  cubes,  but  this 
may  be  done  by  means  of  tetrahedrons  and  octahedrons  which 
have  equal  faces,  by  using  twice  as  many  of  the  fonner  as  of 
the  latter. 

Consider  two  octahedi'a  so  placed  that  two  of  their  edges 
shall  coincide,  and  the  squares  of  which  they  are  sides  be  in 
the  same  plane.  Let  AB  (fig.  87)  be  either  of  these  edges, 
G  a  vertex  of  one  octahedi'on,  not  lying  in  the  plane  of  the 
squares,  D  the  corresponding  vertex  of  the  other.  Then  CD  =  a 
side  of  the  square  =  AB  =  CA  =  CB  =  BD  =  AD^  by  definition 
of  a  regular  octahedron.  Hence  CADB  is  a  regular  tetrahedi'on. 
Hence  if  we  have  a  number  of  octahedra,  so  placed  that  one 
plane  shall  contain  a  square  section  of  each,  and  each  edge  of 
each  such  section  coincide  with  one  edge  of  each  of  the  adjacent 
sections,  an  equal  number  of  tetrahedra  will  fill  up  the  vacant 
space  above  the  plane,  and  therefore  by  using  twice  as  many 
tetraliedi-a  as  octahedi-a,  we  fill  up  the  space  above  and  below. 

6.    Prove  that  the  tangent  plane  at  any  point  of  the  surface 

[axY  +  iJyyY  +  [czY  =  2  {bcyz  +  cazx  +  ahxy)^ 

intersects  the  surface  ayz  +  hzx  +  cxy  —  0  in  two  straight  lines 
at  right  angles  to  one  another. 


1850.]  GEOMETRY   OF   THREE   DIMENSIONS.  21.3 

The  equation 

{axY  +  [hjY  +  {cz)''  =  2  [hcyz  +  cazx  +  ahxy) 

may  be  put  into  the  fomi 

{ax)^  +  [hyY-  +  [czY-  =  0 (1). 

Also  ayz  +  hzx  +  cxy  =  0  may  be  wiitten 

«      ^      c      „  ,_. 

-  +  -4  -  =  0 2  . 

x      y      z 

The  equation  to  the  tangent  plane  to  (1)  at  any  point  [xyz]  is 

Let  [l^m^n^^  iO^h^^)  ^®  ^^^  direction-cosines  of  the  lines  m 
which  (2)  meets  (3),  then  the  condition  of  these  being  at  right 
angles  to  one  another,  is 

IJ^  +  m^m.^  +  n^n^  =  0 (4). 

Now  where  (2)  meets  (3),  we  have,  writing  x^y^z^  for  xyz  in  (2), 

«(ir-(r-(ir-(-)'i(iy^(i)'a' 

a  quadratic  m  --  whose  roots  are  — ^  ,  — ^  .     Hence 


©'^.-(D'^'-e/^-" (^'- 


njWjj        VC2; 


similarly  -^-^  =  f  —  )  : 
n^n^       \czj   ' 

.-.  IJ^  +  ?n,?«,  +  »,?i,  =c  (aa-)*  +  (%)*  +  [czf  =  0  by  (1). 

Hence  the  tangent  plane  at  any  point  of  (1)  cuts  (3)  in  two 
straight  lines  at  right  angles  to  one  another. 

7.  A  certain  territory  is  bomided  by  two  meridian  circles, 
and  by  two  parallels  of  latitude  which  differ  in  longitude  and 
latitude  respectively  by  one  degree,  and  is  known  to  lio  within 
certain  limits  of  latitude  :  find  the  probable  supci*ficial  area. 


214  SOLUTIONS    OF   SENATE-HOUSE    TROBLEMS.  [1851. 

First,  to  find  tlic  ohaiiec  that  tlic  centre  of  the  territory 
which  lies  between  known  limits  a,  /9  of  latitude  lies  in  the  zone 
between  paraHels  of  latitude  J  and  I  +  hi. 

area  of  zone  breadth  hi 


This  chance  = 


area  of  zone  breadth  (a— /3 
27r?-cos/  rhl 


I    'Inrr  cos^  rhl 
cosZ  hi 


(/•  the  radius  of  the  earth,) 


sin  a  —  sin  /3  ' 
Then  the  probable  supei-ficies  of  the  territory 

co%lhl 


j  li      sina  —  sin/S  ' 

A  being  the  area  of  the  territory  when  its  centre  lies  in  the  zone 
between  the  parallels  I  and  I  +  hi; 


1  W+30 

=  — -  {sin  (l  +  30')  -  sin  [l  -  30')}, 
180 


27rr'''sin30' 

cos/. 


180 

Therefore  the  probable  superficies, 
27rr'sin30' 


/3  180(sina  — sm/3) 
7rr'sin30' 


cos' Ihl, 


180(sina-sin/3) 


[a  -  y8  + 1  (sin2a  -  sin2^)}. 


1851. 

1.  A  line  passing  through  a  fixed  point  and  having  the  sum 
of  its  inclinations  to  tAVO  fixed  lines  through  the  same  point 
constant,  generates  a  cone  of  the  second  order. 

Any  section  perpendicular  to  either  of  the  fixed  lines  has  for 
a  focus  its  intersection  with  the  fixed  line. 


1851.]  GEOMETRY   OF   THREE   DIMENSIONS.  215 

(a).  Let  OP  be  the  moving  line,  OS^  OH  the  fixed  lines. 
Take  the  lines  bisecting  the  angle  80H  and  its  interior  angle 
respectively,  as  axes  of  x  and  y.  Describe  a  spherical  surface 
about  0,  cutting  OP,  08,  OH  in  P,  8,  H-,  join  SH,  SP,  HP 
hy  arcs  of  great  circles,  then  by  the  conditions  of  the  problem 

SP  +  HP  =  constant,  2a  suppose. 

Bisect  ^7/  in  X,  join  OX,  let  SX  =  HX  =  ^.  Draw  PM 
an  arc  of  a  great  circle,  perpendicular  to  SH,  let  XM  —  0, 
PM=  (f).     Then,  by  Napier's  rules 

cos  SP  =  cos  8M.C0S  MP, 

=  C0S(/S+  d)  COS(j>, 

COS  HP  =  cos  HM.cos  MP, 

=  cos(/S—  6)  coacf). 

XT  OT>  TTT>         o  SP+HP  SP-HP 

JNow     cos  oP  +  coaiiP  =  2  cos  - — cos ; 

.'.  cosp  cost/  cos(^  =  cos  a  cos , 

.          jrj,              Qp       o    •     SP'rHP   .     SP-HP 
and  cos  HP  —  cos  bP  =  2  sm sm ; 

'   a  '   a       M       '        '    8P-HP 
.'.  smp  sma  cos  9  =  snia  sm , 

therefore  adding  squares 

,, ,  fcoB^  13  cos' 0      sm'8mr^0\  ,  , 

cos'</)    ^^—  + ~, =  1 (1  . 

^  \      cos' a  sm'a      j  ^  ' 

Now  sln^  =  sinPil/  = 


x'-\-y'-\-zy 


2 »        ^  +  y 

.-.    cos  0  =  —, ^ ;  , 

^      x^  +  y^  +  z'^ 

cos  0  =  j-z 2^  7     i^i"  0  =  /  .,      ,M  ; 

{x'+yy  {^'  +  y')^' 

therefore  equation  (1)  becomes 

x' — ^  +  y  ^^  =  x'  +  y'  +  z' 2), 

cos'a       "^  sm'oL  ^  ^  '^ 

shewing  that  tlie  locus  <»f  /'is  a  cone  of  the  second  order. 


216  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1851. 

{j3).   Let  i'cosyS  +  ?ysiii/S  =  ;; (3), 

be  the  equation  of  a  plane  perpendicular  to  OH. 
Where  this  meets  OH,  we  have 

X  =  p  cos^,     y  —  V  sinyS. 
The  distance  of  any  point  [xyz]  in  (3)  from  this  point  is 

If  the  point  {xyz)  also  lie  in  (2),  this  becomes 

fx'  cos'/3  ^  ?/  sin'^/3  _    ,\4 
V  cos"''a      '     sin'''a  /  ' 

or  substituting  for  y  from  (3) 


jiC'^  COs'-'/S        (j3-£CC0S/9f  2I* 

(  cos'^a  sin"''a  J  ' 


which  is  equal  to 


x'  cos''  /3        'Ijix  cos  /3     ^/  cos^  a\  * 


cos  a  sm  a  sm  a  sm  a 


a;  cos/3  cosa 

or  ; p  — —  . 

cosa  sma  sma 

Hence  the  distance  of  any  point  in  the  curve  of  intersection 
of  (2)  and  (3)  from  the  point  of  intersection  of  OH  with  (3)  is 
a  linear  function  of  x,  which  is  a  property  peculiar  to  the  focus. 
Therefore  any  section  pei^pendicular  to  either  of  the  fixed  lines 
has  for  a  focus  its  intersection  with  the  fixed  line.* 

2.  The  locus  of  the  points  in  which  a  prmcipal  plane  of  a 
surface  of  the  second  order  is  intersected  by  the  noi-mals  at  the 
different  points  of  a  plane  section  of  the  sm'face  is  a  conic 
section. 

Let   the  equation  to   the   surface   referred   to  its   prmcipal 

planes,  be  j^-^^  ^  Bf  +  C'/  =  1 (1), 

and  to  the  plane  of  section 

Ix  +  lay  +  nz  =2^ (2)- 

*  Sec  Ilcaru  on  Cunes  of  the  Second  Order,  p.  60,  ct  scqq, 


1851.]  GEOMETRY   OF  THREE   DIMENSIONS.  217 

Then  those  to  the  normal  at  [xyz]  are 

Ax  By  Cz 

Where  this  meets  the  plane  of  yz^  we  have 

i',=3'(i-D,  ^.=^(1-2 

Ay,  Az, 

EUmhiathig  x  between  (1)  and  (2),  and  substituting  the 
above  values  of  y  and  z^  we  shall  obtain  the  equation  to  the 
locus  required,  which  may  easily  be  seen  to  be  of  the  second 
order. 

3.  Noraaals  are  drawn  to  a  surface  at  points  indefinitely 
near  to  and  equidistant  from  a  fixed  point  in  the  surface :  de- 
temiine  and  discuss  the  equation  of  the  surface  generated  by 
the  nomials. 

Take  the  fixed  point  as  origin,  and  the  principal  planes 
through  it  as  planes  of  yz  and  zx.  Let  the  equation  to  the 
surface  be  ^  =  ^^.2  ^  ^y.  ^^^^^ 

The  equations  to  the  noraials  at  (a*,  y^  s),  are 

or  x^-  X  ->r  2Ax{z^-Ax'-  By'-...)  =  0 (1), 

y,-y  +  2By{z^-Ax'-Bf-...)  =  0 (2). 

Again,  since  the  point  (xyz)  is  always  at  the  same  distance 
fi'om  the  origin,  we  have 

X  +  y  +  z  =  rt  , 

or  x'  +  y'+  {Ax'  +  Bf+...Y  =  a' (3). 

The  elimination  of  x,  y  between  (1),  (2),  (3),  would  give  the 
equation  to  the  surface.  But  since  the  point  [xyz]  is  always 
indefinitely  near  to  the  origin,  .r,  y,  z  arc  always  indefinitely 


218  SOLUTIONS   OF   SENxVTE-HOUSE   PROBLEMS.  [1851. 

small,  and  \vc  may  neglect  their  powers  higher  thau  the  second. 
Hence  our  equations  become 

.-r,  —  X  +  2Az^x  =  0, 

y,-y  +  2i?2!,7/  =  0, 

Eliminating  a?,  y  between  these,  we  get 

<      +      y^      -  .2 


(2^2, -1)"^   '   i^lBz^-X)'         ' 
the  required  equation  to  the  surface. 

This  surface  is  evidently  of  the  fourth  order,  and  symmetrical 
with  respect  to  the  planes  of  yz  and  zx.  Its  section,  by  any 
plane  parallel  to  the  plane  of  xy  is  an  ellipse,  which  becomes 
a  circle  when  the  distance  z^  of  the  cutting  plane  from  that  of 

xn  = ^ .     When  z,  =  — r ,  the  equation  becomes  x,  =  0, 

"^      A  +  B  ^      2>1 '  ' 

shewing  that  the  section  is  there  a  straight  line  parallel  to  the 

axis  of  y,  and  similarly  when  ^1  =  ^5  the  section  is  a  straight 

line  parallel  to  the  axis  of  x.  The  points  where  these  lines 
meet  the  axis  of  s,  are  the  centres  of  curvature  of  the  principal 

sections  for  —7  ,  -^  are  the  principal  radii  of  curvature  at  the 

origin.  When  ^1  >  ^  (supposing  A  >  i5),  the  area  of  the 
section  continually  increases,  as  manifestly  ought  to  be  the  case, 
since  the  normals  altogether  diverge  after  ^1  >  ^  • 

4.  A  plane  is  drawn  through  the  axis  of  3/,  such  that  its  trace 
upon  the  plane  of  zx  touches  the  two  circles  in  which  the  plane 
of  zx  meets  the  surface  generated  by  the  revolution  romid  the 
axis  of  z  of  the  circle  [x  —  aY  +  z^  =  c^  (c  <  a)  ;  find  the  equation 
to  the  curve  of  intersection  of  the  plane  and  surface,  and  from 
this  equation  trace  the  curve. 


1851.]  GEOMETRY   OF   THREE   DIMENSIONS.  219 


The  equation  to  the  phinc  will  bo 
z  X 


(0- 


c      [d'-c^f 

The  equation  to  tlie  surface,  generated  by  the  revolution 
round  the  axis  of  z  of  the  given  circle,  is 

\{x'  +  yy--aY^z'  =  6\ 

which  rationalized  becomes 

[x'  +  if  +  z'-^d'-6y  =  lc{'{x'^y'') (2). 

To  obtain  the  curve  of  intersection  of  (1)  and  (2),  we  must 
turn  the  planes  of  ^^  and  xy  round  the  axis  of  _y  till  (1)  coincides 
with  the  plane  of  xy^  and  then  put  z  =  0.     This  is  effected  by 

^^'I'iting  [a^_(^\i.j,-c.^ 

tor  X. 

a 

ex  +  (a^  —  c^)^  z  . 

^ —  for  z. 

a 

or  since  z  is  to  be  put  =  0,  ^^ —  x  for  x.  and  —  for  z :  this 

reduces  the  equation  to 

(a;«  +  f  +  a'  -  6y  =  4  [{d'  -  &)  x'  +  a'y }, 

or  [x'  +  y'  +  d'  -  c'Y  =  -i{a'-  c')  {x'  +  /)  +  4o\y'-', 

which  may  be  reduced  to 

x'  +  y'  -  a^  +  c^  =  ±  2cy, 

or  x'  +  [y±  c)'  =  d% 

shewing  that  the  curve  is  composed  of  two  circles,  the  radius  of 
of  each  of  which  is  a,  and  whose  centres  lie  on  the  axis  of  y,  on 
opposite  sides  of  the  origin,  and  at  a  distance  from  it  =  c. 

5.   Prove  (one  of)  the  two  following  properties : 

(1).  If  [A]^  {D)  be  two  given  spheres  not  intersecting  each 
other,  then  every  sphere  which  cuts  [A]  and  (/>)  in  given  angles 
will  touch  two  fixed  spheres. 


220  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

(2).  If  (/I),  [B]  be  two  given  spheres  cutting  one  another, 
then  every  sphere  which  cuts  {A)  and  {B)  in  given  angles  will 
cut  orthogonally  a  fixed  sphere. 

(I).  Take  the  line  joining  the  centres  of  the  spheres  for  axis 
of  .c,  and  its  middle  point  for  origin :  let  a,  —  a  be  the  abscissae 
of  these  centres,  r,,  r^  the  radii  of  [A)  and  [B] ;  x,  i/j  z  the 
coordinates  of  the  centre  of  a  sphere  which  cuts  {A)  and  {B)  in 
given  angles  a,  /S ;  r  its  radius.  Then  we  must  have 
[x  —  ay  +  /  +  z^  =  r'^  +  1^  —  2rjr  cos  a, 
[x  +  (if  +  if  +  z'  =  r^l;  +  r'  -  2?y  cosyS. 

Let  (&,  0,  0)  be  the  coordinates  of  the  centre  of  a  sphere 
which  this  moveable  sphere  always  touches. 

Adding  and  subtracting  the  above  equations,  we  find 

ic'  +  a'  +  f  +  z'  =  \[r'^  +  r/)  +  r'  -  v{i,\ cosa  +  ?'._, cosyS), 

and  —  Ixb  =  -  -  \rf  —  r^  —  2r  [r^  cosa  —  r^  cos/3)], 

y^  -  d'  =^F  -  a\ 
adding  these  three  equations,  we  have 

{x —ljf-\-  f  +  ^^  =  ?•'  —  r\  r^cos  a  +  )\  cos  /3  +  -  [ii\  cos  a  —  r,  cos  /3)  \  , 

+  i{<^  +  ^-.^  +^  (^-/-O}  +  ^"^  -  «^ (!)• 

It  is  evident  that  the  moveable  sphere  will  touch  the  sphere 
whose  centre  is  at  a  distance  from  the  origin  if  the  right  hand 
member  of  equation  (1)  be  a  perfect  square,  or  if 

jr^cosa  +  r^cosyS  +  -  (r^cosa— r^cos^S)  I  =  2  |rj^+  r./+  -  [rf—  r./)  \ 

a  quadi*atic  for  the  detennination  of  J,  shewing  that  there  are 
two  spheres  which  the  moveable  sphere  always  touches. 

(2).  Also  it  is  evident  that  the  moveable  sphere  will  always 
cut  orthogonally  the  sphere,  the  abscissa  of  whose  centre  is  Z», 
if  the  right-hand  member  of  equation  (1)  assume  the  form 


1851.]  GEOMETRY   OF  THREE   DIMENSIONS.  221 

This  it  will  do  if 

r^  cosa  +  r^  cosyS  +  -  {i\  cosa  —  r^  cos/3)  =  0 ; 

h      »•  COS/8  -  r,  cosa 

or  —  =  ~ ' ' 

a      r^  cosyS  —  r^cosa  ' 

which  dctennines  the  centre  of  the  sphere. 


(     222 


DIFFKIIENTIAT.   EQUATIONS. 

1848. 

cos  37 

Assuming  that  sinx  H is  a  particular   integral  of  the 

equation 

S  +  (-l)^;« (■). 

find  the  complete  integral  of  the  equation 

S-(-l>=^" (^)- 

We  see  by  substitution  that  not  only 

cos  a; 

y  =  smr»  H , 

^  X     ' 

is  a  particular  integral  of  equation  (1),  but  also 

since 

^  X 

Hence  the  complete  solution  of  (2)  is 

.  /  .           cos£c\        -r,  (              sin  a? 
y  =■  A\  sma;  H —  j  +  i?  I  cosa- 

where  A  and  B  are  arbitrary  constants. 

Now  assume  as  the  integral  of  equation  (2), 

.  /  .            cosa-N        „  /              sin.r\ 
y  ~  A\  sma;  ^ J  +  i>  (  cosa; 1  , 

where  A   and  B  are  now  functions   of  x  which   have  to  be 
detennined. 

By  the  usual  assumptions  of  the  method  of  variable  para- 
meters, we  find 

dy        .  f  sina;      cosa?\        -r.  /  .  cosa^      sina;\ 

-^  =  A\  cosa; s—  ]  -  B  [  sma;  H -^-    , 

ax  \  X  X    j  \  X  X    J  ^ 

,  dA  I  .  cosa;\       dB  (  sina;\       ^         ,^. 

and  T-    sma;  H +  -^     cosa; =  0 (3). 

ax   \  X   J       ax  \  X  J  ^  ' 


1849.]  DIFFERENTIAL   EQUATIONS.  223 

.,      dA  (  sin  a?      cos.rN       dB  (  .  cosa;      8111,0? 

Also  ^—     cosa:: 5- ?-     sin^c  -\ ; 

ax   \  X  X    j       ax  \  x  x 

=  ^' (4). 

From  the  equations  (3)  and  (4),  wc  proceed  to  find  -y- ,  -7-  , 

dA   {[              sinicV''      cos  a;  /              siiiicN       /  .           cosoj' 
-5—  \  cosic ^     cosa? 4-    sma?  ^ 

ax   y\  X  j         X      \  X  J       \  x 

sin  a;  /  .  cosicXl 

-  -w  r"^ + — j[  • 

2  f  sina; 

=  X     cosa; 

V  a; 

dA  [^        1        1\         „  /  sina;\ 

or  -7-     14—, 5    =  a?    cosa; , 

dx    \  X'        X  I  \  £C    /  ' 

dA        „  /  sina;\ 

,  „        ,„,   dB            „  /  .           cosa!\ 
and  irom  (3)  —  =.  —  x  \  sina;  4  j ; 

.'.  A  =  a^'^'sinx  —  3/a;  sina%?a;, 

=  a;^  sina?  4-  3a;  cosa;  —  3  sina?  4-  C, 

and  B  =  x^  cosa?  —  3/a;  cosa;^, 

=  x^  cosa;  —  3a;  sina;  —  3  cosa?  4-  D ; 

therefore  the  complete  integral  of  (2)  is 

/  .           cosa;\  ,  „  .  ^  „  .  ^. 

.•.  y  =  [  sma?  4- 1  (a?  sma;  4-  3a;  cosa?  —  3  sina?  4-  C  j, 

/              sina?\  ...  „     ,  „  ^, 

4-  (  cosa? ]  (a?  cosa?  —  3a?  sma?  —  3  cosa?  4-  i/j, 

2       ^J  .           cos.-z;\        ^  /              sina;^ 
=  a;^  4-  6  [  sma?  H \  -^  D{  cosa? j  , 

C  and  D  being  arbitrary  constants. 

1849. 

A  curve  is  defined  by  this  property,  that  the  radius  of  cur- 
vature at  any  point  in  a  given  multiple  of  the  portion  of  the 


224  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

nonnal  intercepted  between  tlic  point  and  tlie  axis  of  abscissa? ; 
prove  that  the  length  of  any  portion  of  tlie  curve  may  be  ex- 
pressed in  finite  terais  of  the  ordinates  of  its  extremities. 

The   lengths   of  the   radius   of  curvature   and   normal   are 
respectively 


dx) 


hence  the  dilFerential  equation  to  the  curve  is 


^"^'^LL  =  „Ji^r^V 


=  ny  \  1  -I- 


d'^y  "^  \  \dxj 

_  A 

dx""  1 

(Fx 

dy'  1 

or  y         I    =  — . 

Lt         — -t,     B'      .^-Ii       (^^W  — 
dy  -  ^^^    '     "   dy  ~\        [dyj  ]  dy ' 

and  cot^  -7-  =  — : 
dy      ny ' 

.".  log  cos^  = ^ogCy ; 

or  cos  0  = 


(GuY 


n 


and  s  =  -^  C^y  "  +  0', 


C,  C  being  arbitrary  constants. 


1850.J  DIFFERENTIAL   EQUATIONS.  225 

lleucc  the  lengtli  of  any  portion  of  tlie  curve  is  known  in 
terms  of  the  ordinates  of  its  extremities. 

1850. 

1.  lff{x  —  a,  y  —  h^  z  —  c)  be  homogeneous  with  respect  to 
X  —  a^  y  —  h^  z  —  Cj  then  y (a;  —  a,  y  —  b,  z  —  c)  =0  is  the  equa- 
tion of  a  cone  whose  vertex  is  (a,  J,  c)  ;  if  the  cone  pass  into 
a  cylinder  by  «,  J,  c  becoming  infinite,  shew  algebraically  that 
the  limiting  fonn  of  the  above  equation  is 

(j)  [nix  +  ny  -^-pz -j-  q^     mx -f  ny  +p'z  +  q)  =  0. 

Let  the  axis  of  the  cylinder,  to  which,  as  its  limiting  form, 
the  cone  tends  as  a,  Z»,  c,  are  indefinitely  increased,  be  parallel 
to  the  intersection  of  the  planes 

mx  +  ny  +  pz  =  0, 

m'x  +  n'y  +  p'z  =  0. 

Then  we  have  ma  +  nb  +  pc  =  a  finite  quantity,  a  suppose, 

m'n  -{  71  b  +  pic  = a' 

Hence  when  «,  Z>,  c,  become  infinite  we  get,  neglecting  a,  a! 
in  comparison  with  a,  b^  c, 

-.^^=^^  =  -.^-^ 0). 

np  —  np      pm  —  pm      mil  —  mn 

Now  since  f  is  a  homogeneous  fimction,  we  have,  if  n  be 
its  degree, 

=  0, 
if      iV         if         if         if         if       ,„> 

Hence  dividing  each  term  of  the  left-hand  member  of  (2) 
by  the  coiTcsponding  member  of  (1),  and  observing  that  when 
a,  6,  c,  become  infinite,  the  right-hand  member  will  vanish  after 
the  division, 

i¥  -  "»  ^  +  (i^»''  -2>"i)  ^  +  (»'i«'  -  »«'«)  '£  =  0...(3) : 

Q 


226  SOLUTIONS   OF   SENATE-HOUSE   PKOBLEMS.  [1850. 

whence,  by  Lagrange's  method,  we  get 

dx  dy  dz 

np  —  i^p      pm  —  p'm      mn  —  m'n  ' 

whence  mdx  +  ndy  +  J)*^^  —  ^j 

m'dx  +  ndy  +  p'dz  =  0 ; 

.".   mx  +  vy  -^  pz  -\-  q   =0, 

m'x  4  ny  +  p'z  +  5''  =  0, 

q^  q  being  constants. 

Therefore  the  integral  of  (3)  is 

(f>  [mx  +  ny  +  p^  +  q,     m'x  +  n'y  +  p'z  •+  q)  =  0, 

the  limiting  fonn  of  f{x  —  a,  y  —  h^  2;  —  c)  =  0,  when  a,  />,  0, 
are  indefinitely  increased. 

2.    Prove  the  following  formulae : 

W-    -  =  '«(^  +  ^')  (1X5:7  + 93113X5 +•■•)• 

cosl(,>-2.-)  «-.l  ^  cs{{n-2r)b-s]  ^     ^^^^^^ 
^  '    sm(a— c>)sm(a— cj...     sm(6— a)  sm(o— c) . . .  ' 

where  s  is  the  smn  of  the  n  quantities  a,  5,  c,...  and  r  is  any 
integer  between  1  and  n  —  \  inclusive. 

(1).   We  have 

Vl.3.5.7  "^9.11.13.15  "^*" 

=  ^{(l?7"3:5)  +  (9j[5"rL13)+- 

=  {(!-})  -  3  (^-i)}  +  {{h-h)  -  3  (,^1-1^)1  +••• 

=  2(j-H^-U..-)-(l  +  ^-^-|+-) (!)• 


1850.]  DIFFERENTIAL   EQUATIONS.  227 

Again,  generally 

1      x'      X*      x''  1   ,      n+x 

l+J  +  -5+T'--=2-x^'^[l- 

Let  ic*  =  —  1,     then  x  =  cos|7r  H —  ^sin^Tr, 
1  +  ic  _  1  +  cos^TT  H —  *  sin^TT 
'  '    1  —  X        1  —  COS^TT  -   —  4  sin^TT 

2  cos'''|7r  H —  *  2  sin^TT  cos^tt 
2  sin^^TT *  2  sin^TT  cos^tt 

,  1        COSiTT  -\ *  siniTT 

■     =  COtiTT     .     f  J 1- 

Sin^TT *  COS  ^77 

=  —  *  COt^TT 

=  £-*^"coti7r; 


^^^  (r=^)    =  -  H-^  +  log  COt^TT, 


and 


2x      2*  +  -  4  2* 

Oj  i  9* 


,      /l+a;\       2* 42*        ,  , 

;  ^^Ir^j  =  4 (-  ^TT  +  log  COt^TT) 


=  ^i  (log  COt^TT  +  ^TT)  -  -i  —    (  log  cot  '^  -  ^TT 

Therefore,  equating  real  and  imaginary  parts, 


1  - 

-l  +  l 

i 

2,2*  (^^^  ^^■^^'^  + 

i^), 

1 

3  ~ 

HA 

- 

...  = 

2.2*  (^"S 

COt^TT 

-i^); 

1    -J-  ' 

1     1 

57 

+ 

...  = 

TT 

2^4' 

AH  ( 

1 

+.. 

•)= 

TT             TT 
2           2.2* 

'   ^^  I1.3.5.7 

TT 

2  (2  +  2*)  ' 

7r  =  96  (2  +  2*)  f— L-  + ^ +. 

^  ^  Vl.3.5.7       9.11.13.15 


228  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

(2).    In  general, 

Let  y  =  wx,  0)  being  one  of  the  imaginary  cube  roots  of  unity, 
then  £""  =  ^  +  -]-  "^TT  "^■" 
Similarly,    e'"'^  =  l  +  ^  +  ^V.., 

^^=^+f+i4+- 

Now  1  +  ft)  +  ft)^  =  0, 

But  e'""  +  £"'•'■  =  e^-^^-**  )^  +  e(-5--*i  )^ 
=  £  -     2  cos  -- 


2  y ' 

1.2.3       1.2..3.4.5.6  ^  V    "r       -r        ; 

=  ^('£"  +  2£-^-'cOS^ 

This  problem  may  also  be  solved  by  putting  the  series  =  w, 
we  shall  then  get  the  differential  equation  -^  —  t«  =  0,  the  in- 
tegration of  which,  when  the  arbitrary  constants  are  properly 
determined,  will  give  the  required  value  of  u. 

(3).   If  r  lie  between  1  and  r?,  we  may  assume 

cos  (?i  +  1  —  2r)  ;r        _         A B 

sin(ic— a)  sin(ic  — 6)...       sin  (a?  — a)       sin(a;  — J)      "*' 

^,  J5,  ...  being  quantities  independent  of  iP, 
.•.  cos(»i+l— 2r)a;=-<4sin(ir— J)sin(a;— c)...+jBsin(a7— a)sin(a;— c)... 

+  . . .  identically.* 


*  We  may  justify  the  above  assumption  by  expanding  both  sides  of  this 
equation  in  terms  of  sinx  and  cosx,  and  dividing  by  cos"''j; ;  the  left-hand  side 


1851.]  DIFFERENTIAL  EQUATlOiNS.  229 

Putting  X  =  rt,  we  get 

cos  {n+  l  —  2r)  a  =  A  ain  {a  —  b)  sin  (a  —  c) . . ., 

^  cos(n+ 1  —  2r)  a  ,  > 

8in(«  — oj  sin(a  — cj... 

t3-    -1    1      7?  cos(w+l-2r)6  ,  . 

bmiilarly  B=   •    /,       x    •   /z.-^ — (2  , 

•^  8in(6-a)  8m(6  — cj... 


cos  («  +  1  —  2r)  a?  cos (n  +  1  —  2r)  a 


sin(ic  — a)  8in(cc  — 5)...       8in(a  — J)  8in(a  — c)...sin(a;  — a) 

cos  (n  +  1  —  2r)  5 
sin  [b  —  a)  sin  (&  —  c) . .  .sin  [x  —  5) 


cos  (n  +  1  —  2r)  a  cos  (n  +  1  —  2r)  & 


'*  sin(a— &)8in(a— c)...sin(a-a;)     8in(5— a)sin(5— c)...sin(6— a;) 

+    .    ^°«(''+!--2'-)-^        ,0 (8). 

8in (aj  —  «)  s,in[x  —  0) . . . 

In  a  similar  manner  it  may  be  shewn  that 

sin  [n  +  1  —  27-)  a  sin  [n  +  l  —  2r)  ^ 


sin(a— &)sin(a— c)...sin(rt— x)     sin(5— a)  8in(5— c)...sin(6— ic) 

,     _    sin  (m  +  1  -  2r)  a;        _^ 

sin(a;  — a)  sin(a;  — J)...         

(3)  coss  +  (4)  sins,  where  s  =  a  +  b  +...+  x  gives 

cos{(n  +  1  -  2r)  a- s}        cos{(«+ 1  —  2r)  &  —  s}  _ 

sm  [a  —  6)  sin  (a  —  c) . . .       sin  {b  —  a)  sin  [b  —  c)...      '"  ^    ' 

the  required  result  proved  for  the  w  +  1  quantities  «,  J,...  a^. 


of  the  equation  becomes /(tanx).(seca:)^''''",  /being  of  n  —  2r  +  1  dimensions, 
or  /(tanx)  (1  +  tan*^;)'''',  which  is  therefore  of  n  —  1  dimensions,  and  the 
equation  becomes  one  of  n  —  I  dimensions  in  tanx ;  it  may  therefore  be 
identically  satisfied  by  the  n  quantities  A,  B . . . .  We  here  suppose  »  +  1  —  2r 
positive:  if  however  2r  >  n  +  1,  we  may  A\Tite  for  2r,  2m  +  2  —  2s,  where 
2a  >  1  <  n  +  1  ;  cos  (n  +  l  -  2r)  x  then  becomes  cos  (n  f  1  —  2s)  x,  in  which 
n  +  1  -  2i  is  always  positive. 


230  SOLUTIONS   OF   HENATE-IIUUSE    PROBLEMS.  [1851. 

3.    Given   f[x)  ^f{y)  =f{x+y)[\  -f[x)f{y)],   find   the 
form  oi  f{;x). 

Since       f[x)  +f{y)  =f{x  +  y)  [1  -f{x)f{y)} (1), 

put  X  =  y  =  0^  then 

2/(o)=/(0){i-7(om; 
therefore  either /(O)  =  0,  or 

1  -/(0)>  =  2, 
giving /(0)=±(-l)i 

Taking  this  latter  value,  and  putting  y  =  0  in  equation  (1), 
f{x)±{-lY-=f{x){l  +  {-l)if{x)}, 

which  gives  f{x)  =  +  (—  1)^  for  all  values  of  x ;  therefore  the 
given  equation  is  satisfied  hj  f{x)  =  ±  (—  1)*. 

Again,  if /"(O)  =  0,  in  equation  (1)  wi'ite  y  =  —  x,  then 
f{x)+f{-x)=f{0){l-f{x)f{-x)], 
.■.f{-x)=-f{x). 
Differentiating  (1)  with  respect  to  y,  considering  x  constant, 

therefore,  putting  y  =  -  x^ 

/(-^)=/(0){i+yW1-^l; 

or  putting  —  x  =  z^ 

/(^)  =/'(o)  {i+TRi'l. 

Now/'(0)  =  some  constant,  C  suppose, 

.-.  ^^  =  a{i+./>)l'^ 

Whence  f{z)  =  tanCs, 

or  f[x)  =  tanCic, 
which  determines  the  foim  of /(ic),  C  being  an  arbitrary  constant. 


1851.]  DIFFERENTIAL   EQUATIONS.  231 

1851. 

1.  Let  P  (fig.  98)  be  any  point  in  a  curve  and  8  a  given 
point  in  a  straight  line  A8\  draw  SU  perpendicular  to  SP^  and 
let  the  tangent  at  P  meet  SU^  8 A  respectively  in  U  and  T\ 
find  the  nature  of  the  curve  when  8U  bears  a  constant  ratio 
to  8T. 

Let  the  curve  be  referred  to  P  as  pole  and  8A  as  prime 
radius;  then 

8U:  8T::  sm8TU:  Bm8UT 

::  8m{d +8PU):  cos 8PU 

: :  sin  ^  +  cos  ^  tan  8PU :  1 ; 

.'.  sin^  +  cos^  tan  8PU=  constant,  e  suppose. 

Now  tan  8PU=  r  -=- , 
dr  ' 

.'.  rcosu  -—  =  e  —  fund, 
ar  ' 

cos^      de  _  1 
e  —  m\d  dr      r  ' 

and  log(e  —  sin  &)  =  log  -  , 

•    a  ^ 

or  sma  =  e  —  . 

r 
Transforming  this  equation  to  rectangular  coordinates, 

y  =  e  (a;'  +  /)*-c, 

or  eV  +  (e'  -  1)  /  -  2ci/  -  c'  =  0 ; 

shewing  that  the  curve  is  an  ellipse  or  hyperbola  according  as 
e  is  >  or  <  1. 

2.  The  equation  c  —  a  cos^  cos0  —  h  ainO  Hin<f)  =  0  may  be 
considered  as  the  complete  integral  of 

dO  # 


{d'  cos'^  d  +  J/  sin"^  d  -  c')  ^  '   [d'  cos'^  ^  +  h'  sin'^  0  -  c^ 


232  SOLUTIONS  OF   SENATE-HOUSE   PROBLEMS.  [1851. 

If  c  —  acos0  COS0  —  b  siii^  sin(^  =  0 (1), 

(a  sin  6  cos  <p-b  cos  6  sin  0)  c?^  +  (a  cos  ^  sin  0  —  ?>  sin  0  cos  0)  (^Z0  =  0, 

and  (1)  may  be  considered  as   the    complete  integral  of  this 
equation,  or  of 

acos^  sin^  —  Jsin^  COS0      asin^  cos^  -  Jcos^  sin^         •••\  n 
c  being  considered  an  arbitrary  constant. 

Now  (acos^sin^— J8in^cos0)'''+c'''  =  (acos^sin^— Jsin^cos^)^ 
+  (acos^cos^  + Jcos^sin<^)'*  by  (1) 
=  a*  cos^  0  +  b^  sm^(f>j 
.'.  a  cos<9  sin^  -  J  sin^  cos^  =  [a^  cos''^^  +  ¥  sm^<f>  -  c'"')* ; 
and  similarly, 

a  sin  6  cos<f>  —  b  cos  ^  sin0  =  (a'''  cos'^^  +  b^  sin^^  —  c''')*, 
therefore  equation  (2)  becomes 


[d^  cos^<^  +  h^  sin"'^  -  c^-       [a  sin  ^  cos  ^  —  b  cos^  sin</>)* 


(     233     ) 


DEFINITE  INTEGRALS. 


1849. 

1.    Shew  that 


^  7,  ax  =  ir. 


+xy 


Putting  X  =  tan''  6,  we  get 


f    ^l[££^  dx  =  4.  r  sin'^^  log  tan^t^^. 
Now  generally  1   f[x)  dx  =  I  f{a  —  x)  dx^ 

•)  0  •'0 

.-.    [     sm'^  log  tsinOdd  =  (  "  cos'^  log  cot^(/^ 

•^  0  •'0 

=  -  1     cos'^logtan^o?^. 


Adding  these  equal  quantities  and  dividing  by  2, 
r^log^  db  =  -  2  [  'co82^  log  tan^  J6>. 

Now,  integrating  by  parts, 
-  /co82^  log  tSinOdO  =  -  ^  sin 2^  log  tan^  +  6  +  G, 

.-.  -  I     co82^  log  tan^c?^  =  i^r, 

r"  a**  logic  , 

2.    Shew  that 

I     log  .r  log  \^-^)  ^-^  =  7r«  (l«^g  ^  -  1) • 


234  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1849. 

Let  I     logic  log  f  — 2 — j  dx  =  w, 

dii  f       lofiTiC 

then  -^  =  2a  \       .,  ^ — r,  dx: 

da  J^  x'  +  a'      ' 


put  X  =  a  tan  0,  then 

dti 


^  =  2  T"  log  (a  tan  ^)  ^7^ (1). 

Now  generally,  I   /(a?)  <;/.c  =1    f{a—x)  dx, 

■Jo  ■^  0 

.♦.  ^  =  2  riog(acot6')rZ^ (2). 


(1)  +  (2)  gives,  dividing  by  2, 

J  /•if 

a  it 


da 


=  2  1      log  a  I 

•'  0 

=  Trloga, 
.*.  u  =  7ra  (loga  —  1)  -I  C. 
And  when  a  =  0,  ?«  =  0, 

.-.  c  =  o, 

and  u  =  ira  (loga  —  1), 
which  was  to  be  shewn. 

3.    Prove  that 

jg  (l  +  2ecos04e)      ^  ^  -^  ~       1  -  e" 

the  upper  or  lower  sign  being  taken  according  as  e  is  less  or 
greater  than  unity. 
We  have 

r  2e  +  (1  +  e")  COS0  ^  _  sinO 

j  (1  +  2e  cos 0  +  e'f       ~  1  +  2ecos0+e' ' 

r  2e+  (1+e')  COS0,        ,^   ,  ^^  _  ,    ,.    .. 

,,   .  ^^ sS— 2N2  loge  (1  +  2e  cos^  +  e')  (?0 

J^  (l+2e  co80  +  e')'     *'    ^  ' 

=  , — 7i n ^z  log  1  +  2e  cos  0  +  c')  +  2e       ,- — j^ sr,  dB 

l  +  2ecos0-fe'     ^^  ^  /,  H-2ecos0  +  ey 


1850.]  DEFINITE    INTEGRALS.  235 

sin  6/  r  cos  9 


=  ...  + 


-_  1   £ 


1  +  2e  cosO  +  e^      j„  1  +  2t;  cost/  +  e' 


I —-^ Ti 7. ,  between  the  limits, 

j„  1  +  2e  COS0  +  e"  ' 


W  7fl   1  + «'  r" 


^0 


+  e'  +  2e  COS0 
I  +  e'  r  dB 


2e     Jo  1  +  e'  +  2e  cos0       2e 
J0 


And 


io    1  + 


e'  +  2e  cos^ 


T.  tan~^ tan^TT  = ^  if  f  <  1 


1  -  <?'  1  +  (.        ^         1  _  e 


tan" 


-  tan^TT  =   g  _     if  e  >  1; 


e'  -  1  e  + 

therefore  the  definite  integral  becomes  in  the  two  cases 

I  +  e'      TT          IT                2e'  ,  2 

+  — :^ —  t •,  —  TT  =  TT  - — 73 5\  and  —  TT 


-      2e      1-e'      2e  2e(l-e")  2e(l-e') 


=  +  TT 


1  -  e 


1850. 

1.    Shew  that 

la  ' 


/  3 

I      tan  X  log  (tan  a?)  dx  =  Jtt'*', 

•' 0 

(;8).  I     -L— — ^  (7a:  =  Tratans  or  Trocots, 


and  that 


sm2£ 
according  as  s  is  <  or  >  ^tt. 
(a).    We  have,  a  being  <  1, 


,x  n-\ 


(See  Gregory's  Examples^  p.  477.) 


236  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

Diftercntiatiug  with  respect  to  a, 


/ 

J  o 


Z"  ,  ,  TT    COSaTT 

Logzaz  =  — 


1  +  z     °  sm  aTT 

Putting  0  =  tannic,  the  limits  of  x  will  be  0,  ^tt,  hence 

.    r*"^     2<.-i    1      /..        \  7             tt'^  cos  aTT 
4  I     tan     X  log  (tana;)  ax  = ^-r, , 

-'  0 

whence,  puttmg  a  =  §, 


27r 


TT   cos 


4  I     tan^a;  logftancc)  £?a:  =  ^       =  ~^  5 


sin- 


/ 


tan  X  log  (tana;)  c?a;  =  ^  tt''*. 


(/8).   Putting  a;  =  a  siu^,  we  have 

^o  {a'-x')i  ^^  _  ^  p-         cos'-'^ 

—  sin^ 


-I—- —  a;  "    -r-— sint/ 

sm2s  sm2£ 

therefore  wTiting  —  6  for  6^ 

=  a  ——————  rfcr. 

-    ^— —  +  sin  0 
smzs 

Adding  these  equals  and  dividing  by  2, 
J-.      «  .  8m2ej_w      1  .  o^ 


—  «  ^     .  ..„ sm 


sin2£  sm  2£ 


=  _^    r       /"i  _  cos- 2s 

sin2£  i_i^  V         (1-  sin"''2£sin-'6')       ' 

Tra  cos'''2e   /'^'^  sec'^^ 


r"  sec 

J_w  1+cos'i 


«     .     „        I         ^r 5^r— ^  f?^. 


8in2e  sin2e  J  i    H-cos''^2£tan*^ 

f  sec^  ^ 

Now  j  1  ^  co8''2£  tan'' 6/  ^^  ""  ^^^^^  tan"' (cos  2£  tan^)  +  C; 

/•'^  sec'-'^ 

•*•    /       ■r~, 27r~z — 27i  ""  =  7rscc2e; 

./_j^l  +  cos'' 2s  tan'^  ' 


1850.]  DEFINITE   INTEGRALS.  237 

r"  {a'-x')i   J  I     1  co32e\ 

.*.    /      ^  ax  =  irax  -r—- ; — — 

J  _„     a  \sm2E       8in2£/ 

sin2£ 

=  Tra  tan  s. 

The  above  investigation   holds  if  s  <  ^tt.     If  e  >  ^tt,   let 
e  =  ^TT  —  e'  (s'  being  <  ^tt),  then 

sin  2s  =  sin2e', 

and   I       '  dx  =  7ra  tans', 

sin  2  s' 

=  ira  cots ; 


.'.    I      1  '    dx  =  TTci  tans  or  ira  cote, 

J-a         « 

^-—  -re 

sin  2  s 

according  as  s  <  or  >  ^tt. 

1851. 

1.    If  3/  be  a  function  of  a:  defined  by  the  equation 

a^"  =  {y  —  nx)"'^^  [y  +  nxf~^^ 

shew  that  f -^  =  r -.-^  = -i^  log^^tJlf. 
}^y  +  l3x      J^/3y  +  n'x      n  +  ^     ^       a 

Since  a'-"'  =  (^  -  na;)"^''  (y  +  nx)"-^ ; 

therefore,  taking  the  logarithmic  differential, 

/        r,\  dy  —  ndx       ,        „>  ^?/  +  ndx 
^  '    y  —  nx  '    y  -\-  nx 

W[n  +  ^)[y^-nx)  +  [n  -  ^){y  -  vx-)]  dy 


2  2     2 

y  —  « ic 


+  n  [[n  —  ^){y  —  nx)  —  (w  +  ^)[y  +  wa;)}  <^] 


y  —  w'^oj 


dx      _        dy       ^ 
'  '  y  +  ^x      /%  +  Ti'^aj ' 

••  J  y  +  ^x~  J  ^y  +  n'x 


238  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

between  proper  limits  of  the  variables.     Now  when  a?  =  0,  y  =  a, 

r      ^^      _  r        dy 
"  Ja  y  +  ^x~  Ja  ^y  +  n'x^ 


d{y  + 


71X) 


^y  +  ri^x  +  w^  +  n^x  ' 

(w  + /8)(^  +  waj)  ' 
y  +  ??a; 


the  required  result. 


w  +  yS 


log 


2.    Detennine  the  value  of  the  definite  integral 

"x'-'{\-xY-'dx 
[x  +  ayp       ' 


f 


Let 


1/ 


X  +  a       1  +  a  ' 
then  when  a;  =  0,  ^  =  0,  and  when  a;  =  1,  ?/  =  1, 

X  = ^ ; 

1  +  a  -  y^ 

.  1  _  ^  =  OL+^Kizl) 

1  +  a  -  y    ^ 

I  +  a 
X  -\-  a  =  a 


t)  =^^'-yy' 


1  +  a-  y^ 

n-x^ 
\x  + 

.    ii-a^r'^{i-yr\^i  +  a-y 

"     [x  +  ay  ofi  1  +  a 

Also    7 r-  =    ,,    ^      ,     , 

{x-VaY      (1+a)"' 
cfe  _  d'y  dy 

X        y        1  +  a  —  ?/ ' 

—  1     +    ^  7 


1851].  DEFINITE   INTEGRALS.  239 

.-.  ^7^^"^l^'  da:  =    ,,/     ,   y'-'  (1  -yf-'dy  ; 

=       1       r(«lI(/3) 

a^(l  +  a)*   r(a  +  /3)  ' 
the  required  value.* 

*  See  Gregory's  Examples,  p.  471. 


(     240     ) 


CALCULUS  OF  FINITE  DIFFERENCES. 

1851. 

1.  PT^  pt^  (fig.  90)  are  two  tangents  to  a  curve  drawn  at 
the  exti'emities  of  any  chord  PS})  passing  through  the  pole  S^ 
and  meeting  a  given  line  SA  in  J",  f,  respectively ;  it  is  required 
to  prove  that  the  cui-ve  in  which  the  sum  of  the  reciprocals  of 
ST  and  St  is  constant  has  for  its  equation 

^  =  1  +  ecos^  +/(sin^)'^, 

where y (sin  ^)'^  denotes  any  rational  function  of  (sin^)'^. 

Let  the  angle  SPT  =  0,  PST  =  6,  SP=r,  then 

r        sin  {0  +  (f)) 
ST^      sin</)       ' 

=  sin^  cot^  +  cos^, 

.    ^1  dr  _ 

=  sm  a  — 77,  +  cos  tf : 
r  da 


1  .    a^^s  a 


putting  We  =  - 


Similarly,  writing  O  +  tt  for  6, 

1  =  -  sin(^  +  7r)  -^  +  Ue^^cos(6'+  tt), 

=  ^m^-^-7^,^^^cos6'; 


1.1         .    ^  /'du(^,^      dn, 


"   ST 


+  s  =  ™^  (11"-^)--^^  ("--"»)' 


=  sin'^-T7^    -^ 


^     /M0+7r-^'e 


de  V     sin  (9 


1850.]  CALCULUS   OF    FINITE    DIFFERENCES.  241 

jNow  by  the  conditions  of  the  problem, 

1  1 

■^™  +  -^  =  rt  constant,  c  suppose ; 


"  dd  \     sin^    j      sin''<9' 

.*.  ?<o^.^  —  ^(Q  =  s'm6  {h  —  c  cot^), 

h  being  an  arbitrary  constant, 

=  b  sin  6  —  c  cos  d, 

2e 
= cos(^  +  a),  changing  the  constants. 

But  by  measuring  6  from  a  proper  point,  we  shall  get  a  =  0, 
so  that  we  may  write 

2e 
^^e+ir  -^e  =  -  —  cos  6*, 

an  equation  of  differences,  which,  when  integrated,  gives 

Uff  =  -  cos^  +  Co, 
a 

C„  being  any  function  of  6  which  does  not  change  its  value 
when  TT  +  ^  is  wa-itten  for  0. 

Therefore  we  may  put 

_l4/(sin^r 
^'  -  a  ' 

and  om'  equation  becomes 

_  1  +ecos^+/(sin^)^ 

"' a  ' 

1 

or,  smce  Ug  =  -  ^ 

^=  1  +  <?cos^+/(sin^)'. 


The  following  Problem  in  Geometry  of  Three  Dimensions 
(set  in  1851)  has  been  omitted. 

Determine  the  surface  generated  by  a  tangent  to  a  right 
cylinder  which  moves  parallel  to  the  base,  and  with  its  point 

s 


242  SOLUTIONS   OF   SENATE-HOUSK   PROBLEMS.  [1851. 

of  contact  lying  on  a  helix  :  shew  also  that  a  hyperbolold  of 
one  sheet  may  be  constnicted  touching  the  cylinder  along  its 
base,  and  such  that  the  required  smface  and  the  h}^erboloid 
are  developable,  the  one  on  the  other. 

(a)  Take  the  axis  of  the  cylinder  as  that  of  z :  let  a  be 
its  radius,  and  let  the  equations  to  the  helix  be 

z  .    z 

X  =  a  cos  - ,     y  =  a  sin  -  . 

Let  I,  77,  ^,  be  the  current  coordinates  of  the  required 
surface,  the  equations  to  the  generating  line  touching  the  cy- 
linder at  (.r,  ?/,  z)  will  be 

P  cos  -  +  77  sin  -  =  a,     t=^z. 
c  c 

Eliminating  z  between  these  equations,  we  get 

t  .    t 

^  cos  -4-7?  sm  -  =  a 
c  c 

as  the  equation  to  the  required  surface. 

(/S)  It  is  obvious  that  the  inclinations  of  a  generating  line 
of  the  cylinder  to  a  tangent  to  the  helix  and  to  a  generating 
line  of  the  h}^erboloid,  are  respectively  constant,  and  that  the 
arbitrary  parameter  of  the  h}^erboloid  may  be  so  detennined 
as  to  make  these  two  angles  equal  to  one  another,  each  equal 
the  angle  t  suppose.  Let  a,  a'  be  two  points  on  the  base  of  the 
cylinder,  indefinitely  near  together,  a/3,  a'/3'  the  generating  lines 
of  the  h}^erboloid  passing  through  them.  Also  let  A,  A'  be 
two  points  on  the  helix,  such  that  the  elementary  arc  AA' 
equals  the  elementary  arc  aa';  and  let  AB^  A'B'  be  the  gene- 
rating lines  of  the  helicoidal  surface  passing  through  A^  A' 
respectively,  and  make  ayS  =  a'/3'  =  AB  =  A'B'. 

We  may  then  shew  that  BB'  =  ySyS',  whatever  he  the  mag- 
nitude of  a(3.  Therefore,  if  the  circle  in  which  the  h}'perboloid 
touches  the  cylinder  be  laid  upon  the  helix  in  which  the  heli- 
coidal surface  touches  it,  the  element  of  surface  between  two 


1851.J  GEOMETRY    OK   THFUCE    DIMENSIONS.  24;{ 

consecutive  generating  lines  of  the  former  figure  may  be  super- 
imposed on  the  con-esponding  element  in  the  latter  figure,  and 
so  for  the  other  elements ;  the  flexure  only  taking  place  round 
the  consecutive  generating  lines.  Hence  the  two  surfaces  are 
developable,  the  one  on  the  other. 

We  proceed  to  shew  that  BB'  =  /3/3',  as  above  stated. 

It  Is  easy  to  see  that  27rc  is  the  distance  between  the  suc- 

c 
cessive  threads   of  the  helix,   and   therefore   -  =  cot  i :    hence 

o 

the  equation  to  the  required  h}'perboloid  is 

■x'  +  f  _  i'  _ , 

It  is  manifest  that  y8,  /3'  lie  upon  the  same  circular  section 
of  the  hyperboloid  ;  the  radius  (?•)  of  this  section,  whose  altitude 
call  2, 

r  =  a\\  -{■  —A    by  the  above  equation 


=  a  (  1  + 

Also  )8yS'  subtends  at  the  centre  of  its  section  the  same  angle 
that  Ota'  does  at  the  centre  of  the  base  of  the  cylinder ; 

.-.  ^^'  =  -  aa 
a 

[        ay8'  sin'^X*      , 

Again,  B  and  B'  lie  on  the  surface  of  a  cylinder  with  the 
same  axis  as  the  proposed,  and  whose  radius  (;•') 

=  [a'  +  AB'^f. 

Also  BB'  is  the  same  part  of  the  tlu'ead  of  a  helix  on  this 
cylinder  that  AA  is  of  the  given  helix ;  the  helices  having 
the  same  distances  27rc  between  the  threads,  and  therefore 
their  lengths  being 

{(27rr')'^  +  (27rc)''')4    and    {(27ra)''  +  (27rcy''}4, 

or     27r  (a''' +  ^^'-^  +  c')*    and    27r(a''  +  c7; 

r2 


244                 SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

=  (  1  H a )  -^-^  J  *•'  c  =  a  cotfc, 


therefore  BB'  =  bb'^  aud  the  surfaces  are  developable,  the  one  on 
the  other.* 


*  We   are  indebted   to  Mjf.  Cayley  for   the  solution  of  the  second  part 
of  this  problem. 


(    245    ) 


STATICS. 

1848. 

1.  A  unifonu  slender  rod  passes  over  the  fixed  point  A 
(fig.  91)  and  under  the  fixed  point  5,  and  Is  kept  at  rest  by 
the  friction  at  the  points  A  and  B:  determine  the  limiting 
positions  of  equilibriiun. 

It  is  evident  that  the  friction  must  always  act  upwards,  and 
the  limiting  position  of  equilibrium  will  be  one  in  which  G 
is  so  near  A,  that  the  friction  is  only  just  able  to  support 
the  resolved  part  of  the  weight  along  the  rod. 

Let  AB  =  a,  AG  =  x,  when  G  is  in  the  limiting  position. 

Resolving  the  forces  on  the  rod  along  it  and  perpendicular 
to  its  length, 

fiB  +  fiB'  =  TFcosa, 

B-    B'=:Wsma; 

and  taking  moments  about  Gj 

B'  {a-\-  x)  =  Bx : 

whence  the  above  equations  become 


Bil =—    =  TFsina. 

V         a  +  xj 


Whence,  by  division. 


a  +  2x 
fi =  cota, 


and   X  =  1^  i 0  ^) 


which  is  the  least  possible  value    for  x:    G  may  be  as  high 
above  it  as  is  consistent  with  the  leaning  of  the  rod  against  B. 


24:i)  JSULL'TIONS    OF    SENATK-Iiursi':    TKoBLliMS.  [1848. 

If  A  were  lower  tluiu  7?,   G  miglit  be  as  low  as  we  please, 
but  at  no  les.s  distance  from  JJ  than  the  above  value  of  x. 

2.  Four  uniform  slender  rods,  AB,  BC,  CZ>,  DA,  (fig.  92), 
rigidly  connected,  form  the  sides  of  a  quadi'i lateral  figm'e,  such 
that  the  angle  ^  is  a  right  angle,  and  the  points  B^  (7,  i>, 
are  equidistant  from  each  other:  when  the  whole  is  suspended 
at  the  angle  vl,  detennine  the  position  of  equilibrimn. 

Let  ^,  y  be  the  coordinates  of  the  centre  of  gravity  of  the 
system  referred  to  AB^  AD  as  coordinate  axes;  and  let 
AB  =  2ff,  AD  =  2^»,  and  the  angle  ABD  =  a ; 

.'.   {2a  +  2b  +  4:{a'  +  h'y\  x 

=  2a.a+2(a''+Z»"04[2a+(a'  +  Z»')4[cos(120°-a)-cos(120°+a)}] 
=  2a"  +  2  {d'  +  hy  [2a  +  («''  4  Irf-  3*  sin  a} 
=  2a'  +  2(«''  +  ^»''')*(2«  +  3i.^'). 
Similarly, 

{2a  +  2Z/  +  4  {pi'  +  W)"^]  y  =  2h'  +  2  [d'  +  h'f  {2b  +  3*a). 
Hence,  if  0  be  the  inclination  of  AB  to  the  vertical, 
^l^  b^-\-{a''  +  by{2h  +  SKa] 
^      a'+  {a^  +  hy{2a  +  SKb)' 

3.  A  string  of  given  length  is  attached  to  the  extremities 
of  the  arms  of  a  straight  lever  without  weight,  and  passes 
round  a  small  pulley  which  supports  a  weight :  find  the  position 
of  equilibrium  in  which  the  lever  is  inclined  to  the  vertical,  and 
prove  that  the  equilibrium  is  unstable. 

The  inclination  of  the  lever  to  the  horizon  will  be  deter- 
mined in  this  case  in  the  method  to  be  shewn  in  the  next 
problem  but  one,  the  point  G  being  now  the  fulcinim,  and  P 
vertically  below  G  instead  of  above  it. 

To  determine  whether  the  equilibrium  is  stable  or  unstable, 
let  the  lever  be  turned  through  a  small  angle ;  then  the  weight 
will  assume  the  lowest  position  it  can,  and  the  normal  at  this 
point  to  the  ellipse  mentioned  in  the  above  problem  will  be 
vertical. 


1848.]  STATICS.  247 

Hence  it  is  evident  that  the  vertical,  through  the  weight 
in  its  displaced  position,  will  intersect  the  lever  on  that  side 
of  the  fulcrum  which  is  lowered  in  the  above  arbitrary  dis- 
placement :  hence  the  system  will  tend  further  from  its  position 
of  rest,  and  the  equilibrium  is  unstable. 

4.  Two  equal  strings,  of  length  ?,  are  attached  to  the  fixed 
points  Aj  B^  and  C,  Z>,  respectively,  which,  if  joined,  would 
form  a  horizontal  rectangle ;  a  sphere,  whose  diameter  equals 
AB^  is  laid  symmetrically  upon  the  strings:  find  the  position 
of  equilibrium  and  the  tension  of  either  string,  supposing 

l>  AC+^irAB. 

Shew  also  how  the  problem  is  to  be  solved  when  this  condition 
is  not  fulfilled. 

The  centre  of  the  sphere  must  lie  in  the  vertical  line  through 
the  point  of  intersection  of  the  diagonals  of  the  parallelogram 
ABCD^  and  each  string  must  lie  wholly  in  the  plane  through 
its  points  of  support  and  the  centre  of  the  sphere. 

Let  AB=2a,  AC=2b^  and  the  depth  of  the  centre  of  the 
sphere  =  z ; 

or     z  =  {i{l-TraY-b'}i (1). 

Let  2'  equal  the  tension  of  either  string ; 

.-.  4:T  jj^ 7^  J  =  weight  of  the  sphere  (2) : 

from  equations  (1)  and  (2)  7"  is  known. 

If  I  =  2b  +  TTff,  0  =  0  and  T  =  cc  ^  the  centre  of  the  sphere 
being  in  the  horizontal  plane  ABCI). 

If  I  <2b  +  ira^  we  must  suppose  the  part  of  the  string  not 
in  contact  with  the  sphere  to  become  rigid,  so  as  to  support 
the  sphere  above  the  horizontal  plane  ABCD.  Each  string 
must  still  lie  wholly  in  the  plane  through  its  points  of  support 
and  the  centre  of  the  sphere. 


248  SOLUTIONS   OF   SKNATE-IIOUSE   PROBLEMS.  [1849. 

1849. 

1.  Two  unequal  weights,  connected  by  a  straight  rod  with- 
out weight,  are  suspended  by  a  string  fastened  at  the  extremities 
of  the  rod,  and  passing  over  a  fixed  point :  detemiine  the  po- 
sition of  equilibrium. 

Let  G  (fig.  93)  be  the  centre  of  gravity  of  W  and  W\  the 
two  weights,  P  the  pulley :  PG  must  be  vertical,  and  bisect  the 
angle  WP]V'. 

Let  TFPTr=  2/,  WW'=2a:  then  PG  is  the  normal  of  the 
ellipse,  which  has  WW  for  foci  and  2l  for  axis-major ;  hence 

WP :  WG  : :  T^PT^  :  WW, 

«^'    ^^'^-wTW''^^ 

W 

and      IV'P  =  jjj; :^,  21 ; 

W+  W      ' 

hence  the  sides  of  the  triangle  WPW  are  known,  and  thence 
its  angles :  therefore  Z  WPG  =  I  WPW  is  known,  and  WGP 
=  TT  —  WPG  —  PWG  is  known,  which  is  the  inclination  of 
TFTF'  to  the  vertical. 

2.  A  smooth  body,  in  the  form  of  a  sphere,  is  divided  Into 
hemispheres,  and  placed  with  the  plane  of  division  vertical  upon 
a  smooth  horizontal  plane :  a  string,  loaded  at  its  extremities 
with  two  equal  weights,  hangs  upon  the  sphere,  passing  over 
its  highest  point,  and  cutting  the  plane  of  division  at  right 
angles :  find  the  least  weight  which  will  preserve  the  equilibrium. 
Determine  whether  the  equilibrium  is  stable  or  unstable. 

Let   a  =  radius  of  the  sphere ; 

X  =  distance  of  the  centre  of  gravity  of  the  hemisphere 
from  the  plane  of  division ; 

W  =  weight  of  the  sphere ; 
to  =  weight  required. 
We  may   consider   the   string    to   become    rigidly    attached 
to  the  sphere  without  disturbing  the  equilibrium :  we  then  have 


1849.]  STATICS.  249 

each  system  of  a  hemisphere  and  weight  attached  prevented 
from  turning  about  the  line  of  intersection  of  the  horizontal 
plane  and  plane  of  division,  by  the  tension  iv,  at  the  highest 
point  of  the  hemisphere. 

Hence,  taking  moments  about  this  line, 

w.2a  =  Wx  -f  ioa\ 


.-.    w  =  -W=lW (1). 

To  consider  whether  the  equilibrium  is  stable  or  unstable. 

If  we  give  either  hemisphere  a  small  angular  displace- 
ment {B)  about  the  above  line,  the  weight  to  rises  through 
a  space  a^,  and  the  centre  of  gravity  of  W  falls  through  a  space 
x.d.  Hence  the  common  centre  of  gravity  of  the  hemisphere 
and  weight  rises  through  a  space 

toae  -  WxO  =  0^   by  (1), 

and  the  equilibrium  is  therefore  neuter. 

3.  A  slightly  elastic  string,  attached  to  two  points  in 
the  circumference  of  the  base  of  a  right  cone,  at  opposite 
extremities  of  a  diameter,  is  just  long  enough  to  reach  over 
the  vertex  without  stretching.  The  cone  is  suspended  by  it 
from  its  middle  point:  find  approximately  the  increase  of  its 
length. 

Let   2?  =  mistretched  length  of  the  string ; 
h  =  height  of  the  cone  ; 
a  =  radius  of  its  base  ; 

z  =  the  depth  through  which  the  cone  falls ; 
2  (?  +  \)  =  the  stretched  length  of  the  string. 

Then,  by  the  principle  that  "  tension  varies  as  extension",  if 
T  be  the  tension  of  the  string, 

T  =  E  J  ,  E  &  constant  weight  ; 


250  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1849. 

h  -\-  z 
and  2  T  j — —  =  W  the  weight  of  the  cone. 

t  -J-  A, 

Also,   (h  +  zY  =  {l  +  \y  +  a'; 

I  I  +  X 

or,  omitting  V  and  the  higher  powers  of  \, 

W         f 


and   A,  = 


2E  {r-dy 


4.  An  equilateral  triangle,  without  weight,  has  three  miequal 
particles  placed  at  its  angular  points;  the  system  is  suspended 
from  a  fixed  point  by  three  equal  strings  at  right  angles  to 
each  other,  and  fastened  to  the  comers  of  the  triangle :  find 
the  inclmation  of  the  plane  of  the  triangle  to  the  horizon. 

Let  J,  y,  2,  be  the  coordinates  of  the  centre  of  gravity  of 
the  three  weights  refen-ed  to  the  strings  as  axes :  ^,  y,  2,  will 
be  subject  to  the  condition 

X  -]ry  -\-  z  =  1^ 

if  I  be  the  length  of  the  strings. 

Let  6  be  the  angle  between  the  nonnal  to  the  plane  of  the 
triangle  and  the  line  joining  the  centre  of  gravity  with  the 
origin,  which  is  vertical ;  this  angle  will  be  the  required  in- 
clination of  the  plane  to    the  horizon.     The  direction-cosines 

of  these  lines  are  -,.  —,.  —,.  and  j=-^ — z=r. — =:^vi ,  T=i — 4 — ^svi  ? 

f^qr^T^^'  respectively; 

l_      ^  +  y  +  2 
••  ''''^''-  ^i'  [x'  +  f  +  zy 

-  1  / 

~  '6^'  {x'+f+zy 


1849.]  STATICS.  251 

5.  A  piece  of  string  is  fastened  at  its  extremities  to  two 
fixed  points :  detennine  from  mechanical  considerations  the  form 
which  must  be  assumed  by  the  string  in  order  that  the  surface 
generated  by  its  revokition  about  the  Ihie  joining  the  fixed 
points  may  be  the  greatest  possible. 

By  Guldinus'  property  of  the  centre  of  gravity,  that  curve 
will  by  its  revokition  generate  the  greatest  surface  whose  centre 
of  gravity  is  furthest  from  the  axis,  i.e.  is  lowest,  when  the 
axis  is  made  horizontal  and  the  plane  of  the  curve  vertical. 
Now  we  know  the  centre  of  gravity  will  assmue  the  lowest 
possible  position  when  the  string  is  in  equilibrium  under  the 
action  of  gravity :  hence  the  curve  required  is  the  common 
catenary. 

6.  It  is  required  to  support  a  smooth  heavy  body,  in  the 
form  of  an  ellipsoid,  in  such  a  manner,  that  a  given  radius 
in  the  body  shall  be  vertical,  by  means  of  supports  at  three 
points :  shew  that  if  /,  ?«,  «,  be  the  direction-cosines  of  the 
radius,  and  the  equation  of  the  ellipsoid 

2  'i  V, 

X       y       z 
{•  - — I —  =  1 

then  the  three   points   in  question  must   be    on    the   curve    of 
intersection  of  the  ellipsoid  with  the  cone 

ll/^  (^.  -  ^)  +  ''^-^  (^  -  ^.)  +  ^nxif  ( i  -  1)  =  0. 

We  will  assmne  the  normals  at  the  three  points  to  meet 
hi  some  point  of  the  vertical  radius.* 

The  equation  to  the  normal  at  x\,  ?/,,  2,,  is 

^  yjL  ?i     ' 

d'  h'  c' 

*  This  is  an  ass'omption :  for  the  ellipsoid  will  be  supported  if  two  of 
the  normals  meet  in  a  point  not  in  the  vertical  radius,  provided  the  resultant 
of  the  corresponding  reactions  meet  the  vertical  radius  in  the  same  pouit 
as  the  third  normal  docs. 


252  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

and  this  line  passes  through  the  points  Ir^  mr^  nr ; 

//•  —  .r,       mr  —  y^       nr  —  z^ 

^  ^7  ii  ^    suppose . 

a^  h'  e 

witli  shnilar  equations  for  the  coordinates  of  the  other  points 
of  support. 

These  equations  may  be  written 

/>•  -  ^  .  =  X,, 
mr  -fs  =  y„ 
nr 7,  s  =  z'. 

whence,  eliminating  r  and  s  by  cross-multiplication, 
fmz^      ny.\  (nx,       Iz\  /7y,       mx\ 

or,  di'opping  the  suffix,  we  have 

as  the  equation  to  the  cone  on  which  the  three  points  of  support 
must  lie. 

1850. 

1.  A  right  cone  is  cut  obliquely,  and  then  placed  with  its 
section  on  a  horizontal  plane :  prove  that,  when  the  angle  of 
the  cone  is  less  than  sin"^^,  there  will  be  two  sections  for  which 
the  equilibrium  is  neutral,  and  for  Intennediate  sections  the 
cone  will  fall  over. 

Let  ABC  (fig.  94)  be  the  section  of  the  cone  through  its 
axis,  by  the  plane  of  the  paper,  to  which  the  cutting  plane 
is  supposed  perpendicular.  Let  the  trace  BP  of  the  cutting 
plane  make  an  angle  6  with  BC:  draw  PD  perpendicular  to 
BP^  and  draw  AEF  through  F,  the  bisection  of  BP. 

Let  2a  be  the  angle  of  the  cone;  then  /.ABP=7r  —  a  —  0, 
BDP  =e  +  OL,  and  APD^^O-a. 


1850.]  STATICS.  253 

Also,  let  AE  =  n.EF^  then 

_AjE  _  APs'mAPE  _  2APsm{e-a) 
"  "  EF  ~  FF  smBFB  ~  BP 

_  2  cos(^  +  a)  slnf^-a) 
~  sm2a  ' 

or   n  sin  2a  =  sin  2^  —  sin  2a; 

.-.  sin 2^  =  («+l)  sin 2a. 

If  ?i  =  3,  E  will  be  the  centre  of  gravity  of  the  part  cut  off, 
which  will  therefore  stand  on  its  base  in  neutral  equilibrium,  and 

sin2^  =  4  sin  2a. 

Hence,  if  sin  2a  <  j^,  there  will  be  two  values  of  6,  each  acute, 
such  that  the  corresponding  cutting  planes  shall  give  neutral 
equilibriiun.     For  intermediate  sections, 

sin  2^  >  4  sin  2a, 

and  therefore   ?i  >  4  ; 

hence  the  centre  of  gravity  will  lie  outside  the  vertical  line  PD, 
and  the  section  will  fall  over. 

2.  The  three  corners  of  a  triangle  are  kept  on  a  circle 
by  three  lings  capable  of  sliding  along  the  circle,  and  the 
circle  is  inclined  to  the  horizon  at  a  given  angle :  find  the 
positions  of  equilibiium. 

It  is  evident  that,  as  the  triangle  is  moved  about,  its  centre 
of  gravity  describes  a  circle  about  the  centre  of  the  circle,  the 
positions  of  equilibrium  are  those  in  which  the  centre  of  gravity 
is  at  the  lowest  and  highest  points  respectively  of  this  circle. 
The  corresponding  positions  of  the  triangle  are  easily  found. 

3.  A  smooth  cylinder  is  supported  in  a  position  of  equi- 
librium by  a  string  which  is  wound  m  times  round  it,  and  then 
has  its  extremities  attached  to  two  points  A  and  B  in  the 
same  horizontal  line.  The  position  of  equilibrium  being  that 
in  which  the  coils  are  separate,  shew  how  it   is  determined. 


254  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

and  how  to  find  the  length  of  the  string  in  contact  with  the 
cylinder. 

Produce  the  straight  lines  of  the  string  to  meet,  as  they 
must  do,  in  a  point :  project  these  produced  parts  and  the 
curved  part  of  the  string  upon  the  axis  of  the  cylinder;  these 
projections  must  be  equal ;  but  the  inclination  to  the  axis  of 
the  straight  and  curved  parts  of  the  string  is  the  same ;  hence 
the  produced  parts  of  the  string  must  be  equal  in  length  to 
the  part  in  contact  with  the  cylinder.  Hence  the  straight 
parts  of  the  string  occupy  the  same  position  as  they  would 
do  if  the  string,  instead  of  supporting  the  cylinder,  ran  under 
an  indefinitely  small  pulley  supporting  a  weight.  This  con- 
sideration determines  the  inclination  of  the  straight  part  of 
the  strmg  to  the  vertical. 

Let  6  =  the  inclination  of  the  axis  of  the  cylinder  to  AB', 

(fi  —  the  inclination  of  the  string  to  the  axis ; 

Q)  =  the  angular  distance  from  the  lowest  generating 
line  of  the  cylinder  of  the  points  where  the  string 
leaves  the  cylinder ; 

2a  =  the  distance  AB- 

2l  =  the  length  of  the  string. 

Then,  if  we  project  the  line  AB  and  the  string  upon  the 
axis  of  the  cylinder,  we  have 

a  cos  0  =  1  cos  cji  ( 1) . 

Again,  if  we  project  AB  and  the  straight  parts  of  the  string 
produced  to  meet  as  above  on  the  plane  of  either  extremity 
of  the  cylinder,  the  line  AB  would  be  projected  into  a  line 
of  length  2a  sin^,  and  the  string  into  two  lines,  each  touching 
the  circular  end  of  the  cylinder,  and  of  length  /sini/r:  and 
these  lines  touching  the  circular  end  of  the  cylinder,  they  make 
with  each  other  the  angle  tt  —  2ft).     Hence 

a  sin  9  ,  . 

coseo  =  ^— ^ — :  (2). 

/  sni9  ^  ' 


185L]  STATICS.  255 

Also  the  produced  parts  of  the  string  each  equals  half  the  part 

m  contact  with  the  cylinder  =  {7mrr  +  &)?■)  cosec<^. 

Hence,  from  the  ahove  projected  triangle, 

(mTrr  +  cor)  cosec  d> 

tan&)  =  ^^ 

r 

=  (mTT+ct))  cosec ^ (3). 

From  (1),  (2),  and  (3),  we  may  determine  6  and  0;  or  the 
position  of  the  axis  of  the  cylinder  and  a :  whence  the  lengtli 
of  the  part  of  the  string  in  contact  is  knoAyn. 

Another  condition  is,  that  the  centre  of  the  cylinder  must 
be  symmetrically  situated  with  respect  to  A  and  B. 

1851. 

1.  A  right  cylinder  upon  an  elliptic  base  (the  semiaxes  of 
which  are  a  and  h)  rests  with  its  axis  horizontal  between 
two  smooth  planes  inclined  at  right  angles  to  each  other :  de- 
termine the  position   of  equilibrium,    (1)   when  the  inclination 

of  one   of  the  planes    is  greater   than  tan"*  -r ,    (2)   when  the 


inclmation  of  both  planes  is  less  than  tan  * 


b 
a 


Since  the  locus  of  intersection  of  tangents  to  an  ellipse  at 
right  angles  to  each  other  is  a  circle,  the  locus  of  the  centre 
of  gravity  of  the  cylinder,  as  the  cylinder  is  turned  about  in 
a  vertical  plane,  is  a  circular  arc ;  and  the  centre  of  gravity 
is  at  the  extremities  of  this  arc  when  the  axes  of  the  cylinder 
are  parallel  to  the  planes.  Also  these  extremities  are  the  lowest 
points  of  the  arc  when  the  inclination  of  both  the  planes  is  less 

than  tan~*  j- ;  but  if  one  of  them  be  greater  than  tan"*  j  ,  one 

extremity  is  the  highest  point  of  the  arc  and  the  other  the 
lowest:  hence,  in  this  case,  the  position  of  equilibrium  is  that 
in  which  the  major  axis  is  parallel  to  the  plane  whose  inclination 
is  least ;  and  in  the  former  case  there  are  two  positions  of  equi- 
librium, viz.  when  each  axis  of  the  cylinder  is  parallel  to  either 
plane. 


256  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

2.  A^  Bj  Cj  are  three  rough  points  iii  a  vertical  plane ; 
P,  Q,  Bj  are  the  greatest  weights  which  can  be  severally  sup- 
ported by  a  weight  PF,  when  connected  with  it  by  strings 
passing  over  A,  B^  C,  over  A^  B^  and  over  B^  C,  respectively: 

shew  that  the  coefficient  of  friction  at  B  =  -  loe;,  -^^r?  • 

We  may  consider  each  of  the  rough  points  A^  B^  C,  as  cy- 
linders of  indefinitely  small  radius :  hence,  by  a  known  theorem 
relating  to  strings  passing  over  rough  smfaces,  if  6  be  the 
angle  through  which  the  string  is  bent  at  any  of  the  points 
whose  coefficient  of  friction  is  ^,  and  T^^  T^  be  the  tensions  of 
the  strings  on  the  two  sides  of  the  point,  if  all  possible  friction 
is  being  exerted,  we  have 

Let  fjbj,^  fjb^,  ficj  be  the  friction  at  A^  B^  and  C;  a,  7,  the 
inclinations  to  the  horizon  of  BC  and  AB  respectively:  then, 
by  the  question, 

p^  £^^(4-7,^gM^(7-;_3Mcli''+').T.]7 (1)^ 

Q  =  e^^Afr-^'.sMBfi^+^J.Tr (2), 

i?  =  £'-,(*'-«). £"c'^'+'MF (3); 

.-.  (2)  X  (3)  -  (1)  gives 


1  ,_    QB 

TT 


/^B=-lOg3p^^. 


(     ^57     ) 


DYNAMICS  OF  A  PARTICLE. 

1848. 

1.  If  a  and  na  be  the  respective  distances  of  a  satellite 
and  of  the  Siin  from  a  planet,  ^  and  mp  the  periodic  times  of 
the  satellite  and  planet,  which  are  supposed  to  describe  circles 
round  the  planet  and  Sim  respectively :  shew  that  the  orbit 
of  the  satellite  will  always  be  concave  towards  the  Sun,  pro- 
vided n  be  greater  than  m^. 

Let  the  angular  velocities  of  the  satellite  and  planet  respec- 
tively in  their  orbits  be  called  eo  and  mw;  then  it  is  plain 
that  the  rectangular  coordinates  of  the  satellite  referred  to  the 
Sun  as  origin  and  axes  rightly  chosen,  are 

X  —  na  coswi  +  a  coswwi, 

y  =  na  sinw^  +  a  sinwio)^. 

Now,  if  the  path  of  the  satellite  pass  at  any  time  t  from 
being  concave  to  convex  towards  the  Sun,  we  have  at  that  time 

M~    '         dt    de       dt    df  ~     ' 

.'.     (n  smoit -'r  m  mima>t)  (n  sinw^-f  ni^  sinmcot) 

+  {n  cosQ>t  +  m  coBmcot)  [n  coBwt  +  nf  cosnicot)  =  0 ; 

.•.  n^  +  m^  +  mn[m  +  1)  cos(7n  —\)(i>t  =  0. 

In  order  that  this  equation  may  not  give  a  possible  value 
of  f,  we  must  have 

n^  +  m^  >  mn  (m  +  1) ; 
.*.  T^  —  m[m-\-\)  n  >  —  ??«"', 


or 


or   w  >  m 


258  SOLUTIONS  or  BENATE-HOUSE  PT^OBLEMS.  [1848. 

which  19  tlicrcfore  tho  condition  to  be  fulfilled,  in  order  that 
tlie  path  of  the  satellite  may  be  always  concave  towards  the 
Sun.* 

2.  A  body  of  given  elasticity  is  projected  with  a  given 
velocity,  and  rebounds  n  times  at  a  horizontal  plane  passing 
througli  the  point  of  projection:  detcnninc  the  direction  of 
projection,  so  that  the  angle  between  the  direction  of  projection 
and  the  direction  of  the  ball  inunediately  after  the  last  impact 
may  be  the  greatest  possible. 

Let  a,  Kj,  a^  •••  ^ni  ^^  ^^^^  angles  of  the  first  projection, 
and  after  the  successive  impacts; 

.'.  tana„  =  e  tana,,_j  =  e^  tana,,,^  =  ...  =  e"tana, 

if  e  is  the  modulus  of  elasticity ; 

,  .       (1  —  e")  tana 

.-.  tan  (a  -  aj  =  ;   ,    J.     ,     : 
"        1  +  e  tan  a 

we  have  to  determine  a,  so  that  this  shall  be  a  maximum. 

Taking  the  logarithmic  diiFerential  of  this  expression  with 
respect  to  tana,  we  have 

1  2e"  tana     _ 

tana      1  +  e"tan''a  ~     ' 

.*.  1  —  e  tan' a  =  0, 
and   tana  =  -r  . 

3.  If  a  body  be  projected  with  a  given  velocity  about  a 

centre  of  force  which  cc  . ,.  ^  ,., ,   shew  that  the  axis-minor  of 

(dist.)^ ' 

*  Since  the  above  condition  assigns  an  inferior  limit  to  the  value  of  m 
(n  remaining  constant),  it  manifestly  precludes  the  possibility  of  a  motion 
of  the  satellite  about  the  Sun  in  a  direction  opposite  to  that  of  the  planet 
i.e.  a  retrograde  motion  as  seen  from  the  Sun,  -which  would  clearly  require 
m  to  be  greater  than  when  its  path  is  merely  alternately  concave  and 
convex  and  not  looped. 


1848.]  DYNAMICS   OF   A    PARTICLt:.  259 

the  orbit  described  will  vary  as  the  perpendicular  from  the 
centre  of  force  upon  the  direction  of  projection;  and  detennine 
the  locus  of  the  centre  of  the  orbit  described. 

Let  r  be  the  distance,  a  the  angle  of  projection : 

then  h^  =  SY.IIZ^  the  product  of  the  perpendiculars  from  the 
foci  on  the  direction  of  projection, 

=  SP  sma.  HP  sma, 
=  r  (2a  — r)  sin'^a. 

And  a  is  constant  since  the  velocity  of  projection  is  so ; 

.'.    b  cc  sina, 

X  r  sin  a, 

GC  the  perpendicular  from  S  upon  the  direction  of  pro- 
jection. 

Also,  if  p,  </)  be  the  polar  coordinates  of  the  centre  of  the 
curve  referred  to  the  centre  of  force  as  pole,  and  initial  radius 
vector  as  prime  radius,  p  =  ae^  <f>  =  angular  distance  of  the 
apse,  c  =  distance  of  projection. 


1 

= 

1 

a 

1 

- 

e  1 

COS(f> 

c 

1 

— 

-7-" 

= 

1 

a 

1 

- 

P 

a 

C0S</) 

1 

X 

or   p^  —  cp  COS0  +  ac  —  a'^  =  0, 
from  which  equation  we  see  that  the  locus  required  is  a  circle. 

4.  Two  bodies,  -4,  B,  when  acted  on  by  gravity,  are  pro- 
jected from  two  given  points  in  the  same  vertical  line  with 
the  same  velocity,  and  in  parallel  directions:  shew  that  if  A 
be  higher  than  B^  a  pair  of  tangents  drawn  to  5's  path  from 
any  point  of  yl's  path,  will  intercept  arcs  described  by  B  in 
equal  times. 

S'2 


260  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

For  if  we  join  the  two  points  of  contact,  the  chord  so  formed 
will  be  an  ordinate  to  the  vertical  diameter  through  the  point 
in  which  the  tangents  meet :  let  2y  be  this  double  ordinate ; 
then,  if  h  be  the  height  of  A  above  J5,  and  the  equation  to 
B'»  path  referred  to  this  diameter  be 

we  have  y^  =  I'h. 

Also,  if  I  be  latus-rectimi  of  the  parabola,  and  a  the  in- 
clination of  the  ordinates  of  the  above  diameter  to  it,  the 
horizontal  distance  between  the  points  of  contact 

=  2y  sin  a, 
=  21'^  sin  a.  A*, 

=  2im, 
is  constant. 

Therefore  also  the  time  of  passage  between  the  points  is 
constant. 

5.    A  body  is  acted  on  by  a  force  =  , ,.     ,^  tending  to  a 

(^dist.j 

fixed  centre  S:  shew  that  in  general  there  will  be  two  direc- 
tions, ditFerently  inclined  to  AS^  in  which  the  body  may  be 
projected  from  a  given  point  A^  with  a  given  velocity  v,  so  as 
to  pass  through  another  given  point  B. 

Prove  also  that  if  t,  t'  be  the  times  of  moving  from  A  to  B 

in  the  two  cases,  either  t  =  t'  or  t  +  t'  =  27r/u.  (  ^  —v^]  *• 

Let  the  body  be  projected  from  A  (fig.  95)  in  a  direction 
making  an  angle  a  with  the  distance,  so  as  to  pass  through  B : 

SA  =  a,   SB  =  b,     lASB  =  ^. 

The  equation  to  the  orbit  is 

d^u  fi        _ 

Tn'2   "1"    "  e    2      '     2        —  ^1 

do               V  a  sm  a        ' 
or   II  =    .^  (1— e  cos(^-7)} (1), 

du  fM  -    fa       \ 

la  =  -2— 2— --2-  .  e  sm  c/  —  7  . 


1848.J  DYNAMICS   OF   A   PARTICLE.  261 

Now,  when  <9  =  0,  u  =  - . 

'  'a 

du  1 

.*•    -  =    .,  ...  .,    (1  —  e  COS7) :     -  cota  =    .;  .^  .  2    e  81117. 
a      vVsin'a^  "^     a  vV  sura 

Eliminating  e  cos 7  and  e  sin 7  from  equation  (1),  we  have 

u  =    ..  ...  ■„ (   .,  .,  .  ■■ )  cos^ cota.sin^; 

V  a  sin  a       \v  a  sin  a      a  J  a 

but  when  ^  =  /9,  ^  =  r ; 

.-.   7  =  -5^,  (1  +  cot" a)  -  \-^,  (1 +cot'''a)  -  -I  cosyQ  -  -  cota  sinyS, 

a  quadratic  equation,  from  which  the  two  values  of  cota  can  be 
detemiined ;  which  proves  that  there  are  in  general  two  diflferent 
directions  of  projection. 

Now  (Hymers'  Ast.^  Art.  326)  the  time  from  A  to  B  can 
be  determined  in  terms  of  the  focal  distances  SA^  SB^  the  chord 
AB^  and  the  axis-major;  and  the  velocity  being  given,  the 
axis-major  is  independent  of  the  direction  of  projection :  hence 
SA^  ABj  BS,  and  the  major-axes  of  the  two  orbits,  are  the 
same.  Therefore  the  periodic  time  in  the  two  orbits  is  the 
same;  and  also  the  time  from  A  to  B. 

If  ^,  t'  be  the  times  of  describing  AB,  and  the  bodies  be 
projected  so  as  both  to  describe  the  angle  ASB^  t  —  t'.  But  if 
one  describes  the  angle  ASB^  and  the  other  27r  —  ASB^  t  +  <' 

equals  the  periodic  time  in  the  conic  section  =  — - —  ,  where 
A  equals  the  semiaxia-major  =  - — — — Tap")  ^^^' 

,  _  ji^    SP{2A-SP) 
"^  ~  SF''  A  ' 

■■■>  +  >■  =  2..  {^-^)-\* 

*  For  this  solution  we  arc  indebted  to  Mr.  Gaskin. 


262  SOLUTIONS  OF  SENATE-HOUSE   PllOBLEMS.  [1848. 

6.    Force   varvinsr   as    y^. — -., .   shew  that,  when  the  hatus- 

•     °  (dist.)  '  ' 

rectum  is  given,  an  angle  =  2  tan"'  5^ ,  measured  from  the  nearer 
apse,  will  be  described  very  nearly  In  the  same  time,  whether 
the  body  moves  ui  an  elliptic  or  an  hyperbolic  orbit,  whose 
eccentricities  are  1  —  a  and  1  +  a  respectively,  a  being  small. 
We  have 

dt  ~7' 

and    r  =  Z  (1  +  (1  +  a)  cos^}"\ 

in  the  ellipse  and  hyperbola  respectively,  where  I  is  the  common 
latus-rectum :  also  h^  =  fxl  is  the  same  in  both  cases ;  hence 

§  =  X 'a +  (!+«)  cos^r 

=  ^-(2cos'i6'  +  aeos^)-'' 

~  h        4       V  cos'^i^j 

r         4  1  /,  /,  cos^  \  , 

=  -y  sec  ho  [1  +  2a  — rr-p^     very  nearly ; 
4 A         ^     \    ~       cos' ^6 J        -^  -^  ' 

.-.   t  =  ^  Jisec^e)  {1  ±  2a  (sec'''i6'  -  2)}  d  tan^^ 

=  ^T  /{I  +  tan'-'i^  ±  2a  (tari*|6'  -  1)}  d  tan  1(9. 

Hence,  if  T  be  the  time  of  describing  an  angle  /3  from  the 
nearer  apse, 

T=^^  [tan  1/3  + 1  tan^iyS  ±  2a  (^  tan^i/3  -  tanlyS)}. 

Hence  the  difference  of  times  of  describing  this  arc  in  the 

two  cases 

2«7^ 
=  f|l  (itan^'iyS-tani^), 

which  vanishes  if /3  =  2  tan~'5%  and  the  proposition  is  true. 


DYNAMICS   OF   A   PARTICLE.  263 

1849. 

1.  If  the  equation  for  detennining  the  apsidal  distances  in 
a  central  orbit  contain  the  factor  {u  —  a)'\  shew  that  a  will  be 
a  root  of  the  equation 

(}>{u)  -  AV  =  0, 

where  ^[u]  is  tlio  central  force. 

The  differential  equation  of  the  orbit  will  be 


nnrl 

dd 


(1). 


Multiply  by  2  -^  ,  and  integrate ; 


(duS"  _      C<f>[u).du        , 


The  general  condition  for  an  apse  is,  that  -7^  =  0 ;  and  there- 
fore the  equation  for  detennining  the  apsidal  distances  is 

If  this  equation  contain  the  factor  [u  —  ay,  let  us  suppose  that 

^/^W^-""=/Wl'.(«-a)'; 

tlien    -^={u-a)  .f{u), 
d'^u      du     d     (du 


^^^  de'~ dd'du'Kde 

=  f{n)[f{u)^f'{u).{a-a)]{u-a)', 
hence  m  =  a  is  a  root  of  the  equation 

dd'  ' 

that  is,  a  root  of  the  equation 

(^[u)  -  //V  =  0. 


264  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1849. 

2.  A  body  moves  from  rest  at  a  distance  a  towards  a  centre 
of  force,  the  force  varj^ing  inversely  as  the  distance :  shew  that 
the  time  of  describing  the  space  between  ySa  and  ^a  will  be  a 

maximum  if  yS  =  — j— . 


We  have  here 


d'^x  _      fx, 
~de~~x' 


{^\  =2/*  log-, 


\dtj  °  03 


smce  a?  =  a,  when  -r-  =  0 : 
'  dt         ' 


_      1  dx 


J  {F{x)  +  C]  suppose. 


Now,  let  T  be  the  time  of  describing  the  space  between 
/8a  and  /S''af  then 

In  order  that  this  may  be  a  maximum,  we  must  have 

d^      ^' 

.-.  n^"-'F'{/3"a)-F'{^a)  =0. 
1 


But  F'{x)  = 


therefore  the  above  condition  becomes 

^^Z i_-o 


log^.) 

or    H*/3"-^-  1  =  0; 


1849.]  DYNAMICS   OF  A   PARTICLE.  265 

.-.-3  =  4:. 

the  required  expression. 

3.  A  particle  is  attached  to  the  extremity  of  a  fine  string, 
which  is  partially  wound  round  a  cylinder  of  diameter  c;  if 
the  unwound  portion  of  the  string  be  kept  stretched,  and  the 
particle  be  projected  perpendicularly  to  its  length  with  a  ve- 
locity F,  prove  that  the  string  will  be  wound  up  after  the  lapse 

P 
of  the  time  -^  ,  where  I  is  the  length  of  string  unwound  at  the 

time  of  projection. 

Let  r  be  the  length  of  the  string  unwound  at  the  time  t 

after  projection,  -  0  the  arc  which  has  become  covered  with 

string  in  that  time:  then,  since  the  only  force  on  the  particle, 
viz.  the  tension  of  the  string,  is  always  pei'pendicular  to  its 
instantaneous  direction  of  motion,  the  velocity  of  the  particle 
is  uniform: 

da  ^  ^ 

.'.   r  -rr  =  a,  constant : 
at 

and  at  the  time  of  projection 

r-=  V' 
dt  ' 

dd       j^ 
dt 

f> 
Also,   r  =  I  —  -6; 

c  ^\d0 


^i^)i=^ 


andi(^-l^)   =C-lVt: 


and  when  <  =  0,  ^  =  0;    .-.   6'  =  ^/'; 


J -'-6]  =P-cVf.. 


206  SOLUTIONS   OK   SENATE-HOUSE   PROBLEMS.  [1850. 

Let  T  bo  the  time  when  the  string  is  all  wound  up ; 
.-.  t  =  T,    when  1  -^d  =  0; 

r 

,       rp 

••  Vc' 

4.  A  particle  describes  an  ellipse  about  a  centre  of  force 
in  the  focus  8  (tig.  96) ;  about  S  as  centre  a  circle  is  described, 
which  'is  cut  by  the  radius  vector  SP  in  the  point  Q ;  from  Q 
a  line  is  drawn  perpendicular  to  the  direction  of  the  particle's 
motion,  which  meets  the  major-axis  in  R:  prove  that  R  is 
constant  in  position,  and  that  QR  is  proportional  to  the  particle's 
velocity  throughout  the  motion. 

From  P  draw  the  normal  PG^  and  from  S  the  perpendicular 
^91^  upon  the  tangent;  also  draw  GL  perpendicular  to  /S'P,  PL 
is  half  the  latus-rectum. 

Now  QR  is  parallel  to  PG ; 

.-.  SR  =  ^.SQ 

^e.SQ, 
by  the  property  of  the  ellipse ;  therefore  8R  is  constant. 

Agam,       QR  =  —  .  8Q  = ^^ .  bQ 

PL.8Q         PL.8Q        1 


~  8PcosP8Y        8Y         8Y' 

and  velocity  cc  -qy-] 

.'.   QR  Qc  velocity. 

Q.  E.  D. 

1850. 

1.  A  heavy  particle  is  fastened  by  two  equal  strings  of 
given  length  to  two  points  in  a  horizontal  line,  and  then  whirled 
round  in  a  vertical  plane ;  the  velocity  is  such  that,  if  one  of 
the  strings  break  when  the  particle  is  either  at  its  lowest  point 
or  half-way  between  its  highest  and  lowest  points,  the  particle 


1850.]  DYNAMICS   OF    A    PAUTICLE.  267 

will  still  continue  to  describe  a  circle:  find  the  least  distance 
between  the  point  to  which  the  strings  are  fastened  that  this 
may  be  possible. 

Let  I  be  the  length  of  either  string,  6  be  the  inclination 
of  the  strings  to  the  horizon  when  the  distance  between  the 
points  is  the  least  possible.  Let  V  be  the  corresponding  ve- 
locity of  the  particle  at  its  lowest  point  before  the  string  breaks ; 
then  [V^  —  2gl &\\\d)^  will  be  its  velocity  when  the  strings  are 
horizontal. 

Now,  if  one  of  the  strings  break  when  the  particle  is  at 
its  lowest  point,  it  will  proceed  in  a  horizontal  circle  about 
the  vertical  line  thi'ough  the  point  of  support  of  the  unbroken 
string,  if  the  velocity  be  such  as  to  produce  a  centrifugal  force 
just  sufficient  to  keep  the  string  at  the  same  inclination  to  the 
horizon,  or,  resolving  the  forces  perpendicular  to  the  length  of 
the  string,  if 

^ ^  .8in0  =  q  cos ^, 

Z  cos  ^  ^  ' 

,cos'^  .  . 

or    l^  =  gl -T—^   (1). 

^    sm^  ^  ' 

If  one  of  the  strings  break  when  the  particle  is  half-way 
between  its  lowest  and  highest  points,  it  will  proceed  to  de- 
scribe a  vertical  circle  about  the  point  of  support  of  the  other 
string,  provided  the  velocity  (F"  — 2^'^  sin^)*  be  great  enough 
to  carry  the  particle  over  the  highest  point  of  the  circle,  i.e.  if 
V""-  2glsm6  >  ^gl. 

Now,  6  is  supposed  to  have  received  its  greatest  possible 
value,  and  therefore,  from  (1),  F  its  least  possible  value ;  hence 

V  -2gl%n\d  =  Sgl, 

or   al  ^^  —  2ql  sin  6  =  3gl: 
•^     smU 

.-.  1  -  sin'^  -  2  sin'^^  =  3  sin^, 

or    sin*  ^  +  sin  ^  =  ^, 

and    8in^  =  -  i  +  (i  +  ^)* 

2 


268 


SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS. 


[1850. 


And  the  least  distance  between  the  points  of  support 
=  2?cos^ 

=  (§)Ml  +  (21)*}W. 

2.  If  P  be  the  perimeter  of  a  closed  curve  described  about 
a  centre  of  force,  t  the  time  of  a  revolution,  h  twice  the  area 
described  in  a  unit  of  time,    and   p  the   radius   of  curvature 


We  have 


at  the  time  <,  prove  that  P  =  h  I    — . 

.  -'or 

p=  rvdt 

J  0 


-/ 
-/i 


dt 

't'de 

V 
dO      , 

r  -7-  .rar 
dr 


P 
rdr 


between  proper  limits. 


[r'-fY 


Now  let  (/-/)i=/(^) 


•'•    /    /  t!_   2u    between  the  above  limits 

=  f{0)d0 

•'  0 

=  /(27r)-/(0)  =  0; 
•    P-  [   P^P 


-I: 


■PJ 
jyr  dr 


dr 
dp 


1850.J  DYNAMICS  OF   A    PARTICLE.  269 

r  —r  dr 
clr 


P 
r'de 


_  rwdd 

~  Jo        P 

•'  (1 


p 

''dt 


3.  If  any  number  of  bodies  be  projected  from  a  given  point 
with  the  same  velocity  in  one  plane,  and  describe  ellipses  round 
a  central  force  which  varies  inversely  as  the  square  of  the  dis- 
tance; find  the  law  of  force  tending  to  the  same  centre,  under 
the  action  of  which  a  body  will  describe  the  curve  which  is 
the  locus  of  the  centres  of  the  different  ellipses. 

Let  /Lt  be  the  absolute  force,  V  the  velocity,  and  c  the  dis- 
tance of  projection.     Then,  if  a  be  the  axis-major, 

i  =  ?-^' is). 

a      c        fj. 
Also  the  equation  to  the  orbit  is 

1  _  1    l-ecos(^-a)^ 
r~  a  1-e'  *' 

and  om'  object  is  to  find  the  relation  between  e  and  a ;  for  if 
Pj  (f>  be  the  polar  coordinates  of  the  centre  of  the  ellipse, 
p  =  ae,     <f)  =  a. 
Now,  when  ^  =  0,  r  =  c; 


1 

c 

1 

or   - 

c 



1 
a 

1 

a 

1  —  e  cosa 

1  —  -  cos<& 

a        ^ 

1-^ 

And  from  (1)  a  is  constant ;  hence  this  equation  shews  that  the 
locus  required  is  a  circle ;  we  may  put  it  in  the  form 

^  -  p  COS0  =  d'  (-  -  -]  (2). 

0  \c       aj  ^  ' 


270  SOLUTIONS   OF  SENATE-HOUSE   PUOBLEMS.  [iBf)]. 

Now,  by  Newton,  Sect.  ii.  Prop.  7,  If  F  be  the  force  in  the 
circle, 

FcrJ-J- 

8F' '  FV ' 
And  fi'om  equation  (2), 

SF.SV=a'c{^--\', 
\a       cj  p 


and   FV  =  p  +  a'ci )  -  : 

\a       cJ  p' 


r  cc  —, 


'  '^^^MHJf' 


oc £ 


y  +  T- 


2— r^ 


C  fl 

1851. 

1.  If  a  body  be  acted  on  by  a  vertical  force  so  as  to 
describe  the  common  catenaiy,  shew  that  the  force  and  velocity 
at  any  point  will  vary  as  the  distance  of  that  point  from  the 
directrix. 

The  equation  to  the  catenary  from  the  directrix,  as  axis  of  x, 
which  we  suppose  horizontal,  is 

X  X 

y  =  \c  [^-  +  e% 
and  the  force  is  wholly  vertical ; 
cFx 


df 

dx 

dt 


and    -Y  =  constant  =  V  suppose. 


1851.]  DYNAMICS   OF   A   I'ARTICLE.  271 

. ,       dy      dy    dx 

Also    -r  =  -#  .  -r 

at      dx    dt 

.".  if  V  =  whole  velocity, 

X  X 

and   V  =  ^V {eP -\- €~ ') 
V 

or  the  velocity  at  any  point  varies  as  the  distance  of  that  point 
from  the  directrix. 


Again, 


d^y  _  d    (dy\  dx 
W  ~dx  \dt)  'di 

ccy; 

or  the  force  at  any  point  varies  as  the  distance  of  tliat  point 
from  the  directrix. 

2.  Force  varj^ing  inversely  as  the  square  of  the  distance, 
a  body  is  projected  from  a  given  point  in  a  direction  making 
an  angle  of  45°  with  the  distance,  and  with  a  velocity  =  ii  times 
the  velocity  in  a  circle  at  the  same  distance :  shew  that  the 
direction  of  the  major-axis  will  be  mialtered  when  the  angle 
of  projection  is  increased  to  cot'\l—n^). 


272  SOLUTIONS   OF  SENATE-nOUSE   PROBLEMS.  [1851. 

The  general  expressions  for  the  elements  of  the  orbit  in 
terms  of  the  distance  (c),  the  velocity  (F),  and  the  angle  (^) 
of  projection,  are 

1  _2  __r 

a      c        /x  ' 
e  cosa  =  — — — —  —  1,    a  the  apsidal  angle, 


,        .           V^c  sin/3  cos/8 
and   e  sma  = ; 


u, 

.'.  cot  a  =  tan/3  —  yvo — = — n 5  • 

V  c  smp  COSyO 

Now,  in  the  present  case, 

V^  =  n^  (velocity  in  a  circle  at  distance  c) 

=  w'  ^  c  =  — ^  ; 
c  c 

.*.  cota  =  tan/S 


=  tan/3 


w''  sin/3  cosyS 

1  +  tan'yS  ^ 
w'tanyS    ' 
2 


.-.  if  y8  =  45°,  cota  =  1 

and  if  /3  =  cot"'(l-w'), 

1         l  +  [i-ny_n'~l-{l-ny_      l  +  l-w''_        2 
l-w"'*      w'''"(l-7i^)~      w'^(l-w''')      "  ^•■'       ~        n'' 

hence  the  apsidal  angle,  and  therefore  the  direction  of  the  axis- 
major,  is  the  same  in  the  two  cases. 

3.  A  body  describes  a  parabola  imder  the  action  of  two 
equal  forces,  one  tending  to  the  focus  and  varying  inversely 
as  the  distance,  the  other  parallel  to  the  axis :  find  the  velocity 
at  any  point  and  the  time  of  moving  between  the  vertex  and 
the  extremity  of  the  latus-rectimi. 

The  resultant  of  the  two  equal  forces  will  bisect  the  angle 
between  them,  and  therefore  be  normal  to  the  parabola :  hence 


1851.]  DYNAMICS   OF   A   PARTICLE.  273 

the  velocity  is  constant;  and  if  p  bo  the  radius  of  cui-vaturc 
at  P, 

-  =  resultant  of  the  two  forces 
P 

=  2  1^  sin  SPY 

Ox 

_2fjL.SY 

~     SJP"   ' 

281^ 
P  =  -gY" ' 

.-.   v'  =  ifl, 

and  V  =  2/A*,  the  required  value. 

Again,  if  S  be  the  length  of  the  arc  from  the  vertex  to  tlie 
extremity  of  the  latus-rectum,  the  time  {T)  of  moving  over  it 

T=2fiiS. 
where   y'^  =  4fe, 


I  +  X      J 
ax 


J  n 


W-TI — ?? 


-  +  a;  +  (/a:  +  x^ 
=  ^Z  log  i ^ +  [Ix  +  .r'^)i 

=  ^Z  log (3 +  2.2*)  +  2*  J  (between  the  limits) 
=  {log(l  +  2i)  +  2i|7; 
r=2/u,*[log(l  +  2i)+2i|  I 


274  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

4.  If  the  product  of  the  velocities  at  two  points  P,  Q  of  the 
parabolic  path  of  a  body  acted  on  by  gravity  be  constant,  shew 
that  the  locus  of  the  pole  of  PQ  is  a  circle  having  the  focus  of 
the  parabola  for  its  centre. 

Let  a,  /9  be  the  angles  between  the  axis  of  the  parabola  and 
SP,  SQ  respectively;  then  the  equations  to  the  tangents  at 
P,  Q  referred  to  S  as  pole  are 

1       2 

-  =  y  {cos^  +  cos(^  — a)}, 

-  =  ^  {cos^  +  cos(^-/3)|. 
r       I 

Hence,  at  the  pole  of  PQ, 

cos(^  — a)  =  cos(^  — y8), 
and  a  does  not  equal  yS ;  therefore 


and  at  the  pole 


2      ' 
1      2  /       a  +  /3  a-yS 

^=Tr^-T-+^'^-2- 

4        a        /8 
=  -^  cos- cos-. 

Now    8P=  I  sec'^, 
4  2  ' 

^g  =  ^sec*^f, 

and  velocity'  at  P  =  2^.;S'P, 

Q  =  2g.SQ', 

.'.  SP.SQ  =  constant  (by  the  question), 

or  sec'^^a  sec'''^/3  =  constant; 

therefore  the  value  of  r  at  the  pole  is  constant,  or  the  locus  of 
that  point  is  a  circle  about  the  focus  as  centre. 


1851.]  DYNAMICS   OF   A    PARTICLE.  275 

5.  Force  varying  as  the  distance,  let  P,  Q  (fig.  97)  be  two 
points  in  the  orbit  described  by  a  body  round  a  given  centre  of 
force  C,  and  let  PT  (the  tangent  at  the  point  P)  meet  CQ 
produced  in  T;  join  PQ  and  draw  TA  parallel  to  PQ  meeting 
CP  produced  in  A  ;  draw  QA  meeting  PT  in  Z/,  and  CU 
meeting  PQ  in  F;  in  CUtake  CB  a  mean  proportional  between 
CV  and  CU,  then  the  body  will  pass  through  the  point  P,  and 
the  time  of  movmg  from  P  to  B  will  be  half  the  time  of  moving 
from  P  to  Q. 

Let  CP=a,  CQ  =  h,  then  the  equation  to  the  ellipse  referred 
to  CP,  CQ  a,s  axes  wnll  be 


+  -3^  +  f.=  i (1) 


Let  CA  =  a\  CT  =  h' ;  the  equation  to  PT  will  be 


X 


y 


-+rr=l (2). 


a 


To  express  the  condition  in  order  that  this  may  touch  the 
ellipse,  take  (1)  —  (2)"'' ;  therefore 

an  equation  to  be  satisfied  only  by  the  coordinates  x  =  a,  y  =  0  ; 

therefore 

1         1 


7'         ^ 

or  0  =  -  . 
a 

Also,  since  ^  Z"  is  parallel  to  PQ, 

a    J,       c'' 

which  is  evidently  the  condition  in  order  that   QA  may  touch 
the  ellipse. 

Hence  QU,  PU  both  touch  the  ellipse,  and  by  a  known  pro- 
perty of  the  ellipse,  P  is  a  point  in  the  curve  if  CB'  =  CU.  CV. 

t2 


276  b(JLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

Again,  let  ^,,  </>,  9„  be  the  angles  C'P,  CJR^  CQ  respectively 
make  \A\\\  the  apsidal  distance,  then  the  time  in  which  the 
particle  will  reach  P  from  the  apse 

=  ltan-(|tan^,), 

where  a  and  5  are  now  the  semiaxes. 

Let  (/<,,  /t,),  (^.^,  l\)  be  the  coordinates  of  the  points  P,  Q  re- 
ferred to  the  axes  of  the  figure,  the  equations  to  PZ7,  Q  U  will  be 

K         K 

and  -^,x  +  -^,y=\, 
and  the  equation  to  CZ7  through  their  point  of  intersection 

and  this  line  passes  through  the  origin ;  therefore 

\  =  -  1, 
and  the  equation  to  C?7  becomes 

a"- 


therefore  time  from  PXo  R 


=  -i  |ta»"'  (j  tan<^)  -tan-^  (f  *^^^i) 
_  1      a       tan^  —  tan^, 

1  +  r?  tant^,  tan© 
6  ' 


1851.]  DYNAMICS   OF   A   TAKTICLE.  277 


DYNAMICS   OF   A 

rAKTICLE, 

K 

K  -  K 

a^ 

«^ 

" 

h  -  K 

ah 

h' 

?/ 

/** 

/^(/^-/^J 

1"  +  , 

a            rt* 

'    /^.(^\-^J 

6* 

^1^2  -  ^l^a 

ah 

av;' 

^4 

">0 

1 

^A    \K 

^4., 

ah' 

» 

1 

d'      h' 

an  expression  which  only  changes  sign  when  the  suffixes  1  and  2 
are  interchanged ;  it  therefore 

=  -J  jtan-^  [^  tan  6',^  -  tan"^  [^  tan<^^ 

=  time  from  B  to  Q. 

Q.  E.  D. 

6.  Two  bodies  A  and  5  revolve  in  the  same  conic  section 
roimd  the  same  centre  of  force  in  the  focus ;  shew  that  if  A  and 
B  be  at  opposite  extremities  of  any  focal  chord,  B  will  appear 
(to  a  spectator  on  A)  to  move  with  a  constant  velocity  pei'pen- 
dicular  to  SP^  or  with  a  constant  velocity  perpendicular  to  the 
transverse  axis,  according  as  A  and  B  describe  the  conic  section 
in  the  same  or  opposite  directions. 

Let  PT,  QT  (fig.  98)  be  the  tangents  at  the  points  P  and  Q 
at  the  extremity  of  the  focal  chord  BSQ'j  draw  Sl\  SY'  per- 
pendiculars from  S  upon  those  tangents. 

First  J  suppose  the  bodies  A  and  B  to  be  moving  in  the  same 
direction  about  S^  then  the  relative  velocity  of  A  and  B  per- 
pendicular to  FSQ 

=  l\smSPV+  V^^mSQY', 


278  SOLUTIONS   OF   SENATE-HOUSE    FHUBLEMS.  [1851. 

it'  \\  aiul  r,  be  the  velocities  of  Ji  and  B 

_J_SY        h     SY' 
~  SY'  SP'^  SY'  SF 

~     \SP^  SF 

h 

=  ,  ,7 : IS  constant. 

^  lat.  rectum 

Secondly^  suppose  A  and  B  to  be  moving  in  opposite  direc- 
tions about  S^  then  their  relative  velocity  pei'pendicular  to  the 
transverse  axis 

=  \\co^PTS-  V^cosQT'S 

-  cosSPY-  oD>    -^^  on^^' '  -  cosSQY' 


SP  sin  SPY '  e  SF  sin  SQ,  Y' '  e 


=  -  I  —  cotSFY--^  cotSQY''] 


~e  [sF 


=  -  {u  cot<^  —  u'  cot0')  suppose. 


Let  ASP  =6,  then 

2 

7 


2 

If  =  -  (1  +e  cos^)  [I  the  latus-rectmu), 


A  ^A.  \    du 

and.  cot  m  = ^^ : 

^  u  do' 

du      2e    . 
.•.  i(  cot 9  =  — Jn—  T  sine', 

and  writing  tt  +  ^  for  ^,  and  neglecting  the  change  of  sign, 

since  cotd)'  =  —  ^77  and  not ,  -j^  as  above,  we  find 

^       u   do  u    do  ' 

?i  cot  9  =  -y  sma, 

whence  we  see  that  the   above   expression   gives  the  relative 
velocity  perpendicular  to  the  transverse  axis  equal  to  zero. 


1851.]  DYNAMICS   OF   A   PARTICLE.  279 

7.    A  body  m  moves  with  a  uniform  velocity  v  =  — —  /i* 

in  a  circular  tube  whose  radius  is  a,  and  attracts   a  body  m' 

within  the  same  tube  with   a  force  =  , ,.     ,„;  shew  that  if  m 

and  m  be  originally  situated  in  the  opposite  extremities  of 
a  diameter  and  m  at  rest,  the  two  bodies  will  meet  one  another 
2aa 


at  the  end  of  the  time 


b  sin  a  * 


Let  P,  Q  (fig.  99)  be  the  positions  of  m  and  m  at  the  time  t 
after  the  beginning  of  motion,  and  let  PCA,  QCB  be  their 
angular  distances  from  their  original  positions  ^,  J5  at  the 
extremities  of  the  diameter  AGB\  let  PGA  =  6^   PCQ  =  (f)y 


V 


also  QCB  =  -  t:  join  PQ ;  then,  for  the  motion  of  m', 


a 


fl      cos|0 


(2a)'  mi^^<f>' 


V 


Also,    d  =  TT  —  (b t ; 

a 

•*•   de  "      df  ' 

and    2  '-^  ^  =  _  iL  .^^^  ^ . 
dt     df  4«*  3in'^<^   dt  ' 


dt  J        4a*  Vsin*^)^ 
V*  cos'^'a 


sin''^<^ 


+  C 


Now,  when  <  =  0,  -^  = ,    d)  =  7r: 

'  dt  a 

v^      v"  cos'^a  ,,      ^, 

•••  a' — ^('+^); 

.-.    C  =  SGC'OL-  1, 


280  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

(It  a  b1ii^(/) 


d<f)  V  cosa  [1  +  sijr^^(sec'''a  -  1)}* 


V  cosa  {sec" a  -  (sec'"' a  —  1)  cos''^^^]* ' 

,  2a  1  .  _,  (scc'^a— 1)*        ,  ,       ^ 

and  f  = 7 — r, -Ti  sin    -^ —  cos*©  +  6 

I' cosa  (seca— 1)*  sec  a  ^^ 

2a 

—  — : —  sin  ^  sina  cosi6  +  (7(=0). 
V  sin  a 

And  if  T  be  the  time  that  elapses  before  the  collision, 

J.  =  — ; —  Sin     sma 

V  sma 

2aa 


V  sm  a 


8.  A  straight  rod  AB  (100)  slides  between  two  planes 
0-4,  (9J9,  one  of  which  is  horizontal  and  the  other  vertical : 
then,  if  a  body  acted  on  by  gravity  descends  from  rest  from 
the  highest  point,  down  the  curve  which  always  touches  AB^ 
the  time  down  any  arc  :  the  time  down  the  corresponding 
chord  : :  twice  the  arc  :  the  chord. 

Let  the  length  of  the  rod  AB  =  c,  the  equation  to  the 
cui've  which  always  touches  AB^  referred  to  Cx^  Cy^  the  ho- 
rizontal and  vertical  axes  through  its  highest  point,  is 

(c  —  xy  -f  y^  =  c\ 

We  will  shew  that  CD  is  the  curve  which  has  the  required 
property.     Let  a;',  y'  be  the  coordinates  of  the  point  P.     Then 

time  down   arc  CP  —  '   -^- 


•'  0 


chord  CP  =  y-^-^^  ,  7  the  length  of  the  chord. 


1^51.]  DYNAMICS  OF   A   PARTICLE.  281 

Hence  the  property  in  the  question  gives 
ds 

^Fi "l ' 


ds 

.    f  dy     .  4      tUs  , 

•       1      ^  ^  _£__  ^  _    2     r"'ds 
"   {gy'Y  dy       [gyy  dy      ^^A^^' 

.    _3_   ^^_2_   £   r^V7.9 
"  WJf  dy'      [gyy-y'  Idy'^y^ 

3    ,  ds        ["'ds  ^ 

"~^'  w^Ldy^y^ 

.    3    ,  ^'      3  ^       ds_ 
"  2  ^   dy   ^  2dy'"d^''' 
or  dropping  the  accents, 

ori  +  3^'  =  0, 
y         ds        ' 

log.y  +  3log^  =  logc; 

•••  r  =  (-)'' 

»-(iy=(-;)' 


dy  \      ' 


282  SULUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1851. 

.-.  x-{-  C=-  {c^-ff, 
and  the  curve  passes  through  the  origin ; 

.-.   0=-c, 

i  s  s 

and    [x  —  cy  +  ^^  =  c' 

is  its  equation,  shewing  that  the  curve  which  has  the  required 
property  is  the  curve  generated  as  CD  is. 


(     283     ) 


RIGID  DYNAMICS. 

1848. 

1.  A  GIVEN  Inelastic  mass  is  let  fall  from  a  given  height 
on  one  scale  of  a  balance,  and  two  inelastic  masses  arc  let  fall 
from  different  heights  on  the  other  scale,  so  that  the  three 
impacts  take  place  simultaneously :  find  the  relations  between 
the  masses  and  heights  in  order  that  the  balance  may  remain 
permanently  at  rest. 

Let  M  be  the  given  mass,  h  the  height  from  which  it  falls ; 
iJ/,  M,^  the  other  two  masses,  A^,  h^  the  heights  from  which 
they  fall:  then  the  momenta  of  the  three  will  be 

M{2gh)^^   M^[2g\)^^  M,^[2gh^)^  respectively: 

in  order  that  equilibrium  may  not  be  disturbed,  we  must  have 

sum  of  momenta  of  iHfj,  M^  =  momentum  of  M^ 

or   M^h^i  +  M^^  =  JfA* (1). 

Also,  in  order  that  the  balance  may  remain  permanently  at 
rest,  we  must  have 

M^^M^  =  M  (2): 

(1)  and  (2)  are  the  required  relations, 

2.  A  cannon-ball  is  fired  at  a  mark  at  a  place  whose  north 
latitude  is  Z;  shew  that  in  consequence  of  the  Earth's  rotation 
the  vertical  plane  containing  the  axis  of  the  cannon  must  be 
inclined  at  an  angle  of  \ht  sin/  seconds  to  the  left  of  the  vertical 
plane  passing  through  the  mark,  t  being  the  time  of  flight  in 
seconds. 

The  Earth's  motion  (oj)  of  rotation  about  its  axis  of  figure 
may  be  resolved   into  two ;   one  about  the  vertical  line  at  the 


284  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

place  In  question,  and  another  about  an  axis  tlirough  the  Eaith's 
centre  at  right  angles  to  tlie  former.  The  latter  rotation  will 
not  affect  the  relative  position  of  the  cannon  and  mark. 

The  former  velocity  of  rotation  is  w  sinZ,  and  carries  the 
mark  round  the  camion  from  right  to  left:  consequently  the 
vertical  plane  through  the  cannon  and  mark  will  in  time  t 
revolve  through  an  angle  to  sin  ?.^,  or  faWtlo  ^^^^^'^  degrees,  if 
t  be  the  number  of  seconds  in  the  time  of  flight,  or  through 
an  angle  Yf  sin/.f,  or  15  sin 7.^  seconds:  in  order,  therefore,  that 
the  ball  may  hit  the  mark,  it  must  be  aimed  15  smLt  seconds 
to  the  left  of  the  mark. 

3.  An  imperfectly  elastic  homogeneous  rough  sphere  is  pro- 
jected obliquely,  without  rotation,  against  a  fixed  plane ;  if  t',  i' 
be  the  angles  of  incidence  and  reflexion.  A,  the  coefficient  of 
elasticity  for  direct  impact,  and  p  the  ratio  of  the  tangential 
force  of  restitution  and  compression,  prove  that 

2|0  =  5  —  7A,  tani'  coii. 

Let  Rj  jR,  be  the  normal  Impulses  up  to  the  time  of  greatest 
compression  and  during  the  whole  impact  respectively  : 

Fj  F^  the  same  tangential  impulses, 

F,  V  the  velocities  of  the  centre  of  the  sphere  before  and 
after  impact; 

.'.  R  =  Vcost.Mj 

and   V  co3^'  =  "t?  —  Fcos«  =  XFcosz  (1). 

Also  tangential  velocity  before  impact  =  Fsin/;  and  at  the 

time  of  greatest  tangential  compression  the  tangential  action  F 

.       F 
has  diminished  the  velocity  F sin*  by  the  quantity  -r^,   and  has 

generated  an  angular  velocity  w  where 

Mk'rj  =  Fa. 


1848.]  RIGID   DYNAMICS.  285 

Wc  must  also  express  the  geometrical  condition   that  the 
point  in  contact  with  the  plane  is  at  rest,  or 

.    .      F 
Fsmi  —  ^-p=  a-a. 
M 

„  rr  .    .       F       Fd' 

Hence    K  smi  —  ^r^  =  ^rryr, ; 

.-.  F=    ,^\,  J/Fsln?'. 

af  +  h' 

Also   F^  =  {\-^p)F; 

.    .     F 

.'.   F'sint"  =  Fsini  —  ~ 

M 


=  Fsm;{l-(l+p)^3,} 

(2): 


Tr  .    .a  —  pk 
=  V  sni  I  —T, — St 

ft'  +  h"" 

(2)  4-  (1)  gives 

.,      tan^  a^  —  pF 

tan^  =  — r t. — ^  : 

X      ft'  +  Z;'  ' 

or,  substituting  §ft''  for  A;', 

.,      tan  I    5  —  2p 

.*.  2p  =  5  —  7X  tan 2*'  cot?'. 

4.  Two  given  masses  are  connected  by  a  slightly  elastic 
string,  and  projected  so  as  to  whirl  round:  find  the  time  of 
a  small  oscillation  in  the  length  of  the  string.  Give  a  nu- 
merical result,  supposing  the  masses  to  weigh  1  lb.,  2  lbs.  re- 
spectively, and  the  natural  length  of  the  string  to  be  1  yard, 
and  supposing  that  it  stretches  ^  inch  for  a  tension  of  1  lb. 

The  tension,  and  therefore  the  extension,  of  the  string  will 
evidently  depend  only  upon  the  relative  motion  of  the  masses, 
not  upon  their  absolute  motions.  Now  the  relative  motion  of 
the  masses  will  not  be  affected  if  we  apply  at  each  instant  to 


=  the  extended  length  of  the  string  at  tune  f, 


286  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1848. 

T 

both  bodies  [M  and  M')  the  accelerating  force  vf,  equal  to  that 

acting  upon  il/,  and  in  the  opposite  direction  :  and  if,  further, 
wc  apply  at  the  instant  after  projection  to  both  bodies  the  same 
velocity,  viz.  a  velocity  equal  to  ilf 's  velocity  of  projection  in 
the  opposite  direction  to  it.  But  by  these  means  J/  is  reduced 
to  rest :  let  it  be  taken  as  the  pole  of  coordinates.     Then,  if 

—  =  the  uncxtcnded  length  of  the  string, 

1 
u 

and  T  =  tension  at  time  ^, 

we  have,  by  the  principle  that  '  Tension  ex  Extension', 

1  _  I 

T=E  ''t !i  =  E "^IsiIL:^     e  a  constant  weight. 

1  M  " 

Now  the  equation  of  il/'s  motion  is 

Let  u  =  u^  —  a,  a  will  be  very  small, 
and  ^-,^„  +  a  +  ^  _  +  _  1  +  3=0; 


dd-"     «  '  "  ^  ^  Viif  ^  M'j  7/v;  V       u^ 

or,  omitting  d\ 

which  equation  shews   that  a  undergoes  periodic  mequalities, 
whose  period 


E_(l_       l^ 


l+T^i^  + 


1848.]  RIGID   DYNAMICS.  287 

this  is  the  time  of  the  small  oscillations  in  the  length  of  the 
string :  h  must  be  determined  by  the  circmnstances  of  projection. 

Ex.    Let  M=  1  lb.,  M'  =  2  lb.,  -  =  1  yard,  and  E  such  a 
weight  that  a  tension  of  1  lb.  stretches  the  string  ^  inch,  or 

iib.  =  ^Li^^«i^  =  j?- 

or^=      ^^Ibs.; 
.'.   y  =  27r  .  ,       ^  ,..   ^     1  seconds 


=  ^'^  /^^32,2   x\    360    3^^  '''"^^'' 


(' 


v^        '  359  *  2, 

where  v  is  the  velocity  of  projection,  expressed  in  yards,  of  M 
or  M'  in  their  relative  orbits, 

seconds  very  nearly. 


" 145\i 


5.  A  rough  sphere  rolls  within  a  hollow  cylinder  with  its 
axis  vertical,  so  as  to  be  in  contact  with  the  cm-ved  smfacc 
and  the  flat  bottom:  find  the  reactions  and  the  fiictions,  in 
terms  of  the  angular  velocity  with  which  the  sphere  goes 
round,  and  explain  the  indetenninateuess  of  the  problem. 

Let  (o  be  the  angular  velocity  of  the  centre  of  the  sphere 
about  the  axis  of  the  cylinder; 

o).,  (u_^,  co^  the  angular  velocities  of  the  sphere  about  that 
axis  and  two  other  axes  of  rectangular  coordinates ; 

B^^  i?^,  B^  and  i?'^,  i?'^,  B\  the  mutual  actions  at  the 
base  and  other  points  of  contact  parallel  to  the  axes 
of  Xj  y,  and  z  respectively ; 

a  and  r  +  a  the  radii  of  the  sphere  and  cylinder. 


288  SOLUTIONS   OP  SENATE-HOUSE   PROBLEMS.  [1848. 

Thcu  the  equations  of  motion  arc 

M"^  =  i?^  +  n\  -  Mg  =  0, 
MU'^  =  -  Ra-Ea^mO 

(it  1/  e  , 

[6  the  angle  between  the  radius  vector  and  axis  of  x,) 
JfF^^  =  i2a  +  -B'acos^, 

MT^  ^'  =  Ea  sin<9  -  E  a  cos(9. 
dt 

The  geometrical  condition  to  be  expressed  is,  that  the  two 
points  of  contact  must  be  instantaneously  at  rest. 

Hence, 

horizontal  motion  of  the  point  of  contact  of  the  curve  surfaces,  or 

aa>^  —  r«  =  0 ; 

vertical  motion  of  the  same,  or 

ao)^  sin  6  —  aco^  cos  ^  =  0 ; 

and  horizontal  motion  of  the  other  point,  or 

aco^  cos  6  +  aa>^  sin  0  —  rco  =  0. 

Hence, 

aco^  =  rco  cos^, 

ao)^  =  rco  sin^; 

.'.   a  -7-  =  r  cosd.f—  r  sin^.w", 
dco 


(where  /=-£), 


1848.]  RIGID   DYNAMICS.  289 

a  —jJi  =  r  {iiT\u.f+  r  cosc/.eu  , 


dt 
It 


dto 


Also, 

X  =  r  cos^, 

y  =  r  sin^; 

d^'x 
.".   -j^  =  —  r  sin  O.f—  r  cos6.q)'\ 

— ^  =      r  cos  0.f  —  r  sill  6.q)\ 

Hence  the  equations  of  motion  become 

-  Mr  sm0.f-  Mr  cos^.w'  =  R,  +  R\ (1), 

Mr  cose.f-  Mr  sin^.o)'  =  E  +  R\ (2), 

0  =  i?  +  ir,  -  Mg (3), 

MFr  cos e.f-  Mh\  sin  O.ui'  =  -  Ra,'  -  R! p,'  sin ^ (4), 

MWr  sin  B.f  +  MU'r  cos  6. ai'  =      R/i'  +  R'ci'  cos  6   (5) , 

Mk'rf  =   R'/i'  sin  0  -  R\d'  cos  ^  ...  (6), 

(4)  cos^  +  (5)  sln^  +  (6), 

2Mk\f=:  [R^  +  R',)  d'  sin  6*  -  [R^  +  JS;)  a^  cos  (9 
=  -Ma'rf,     by  (1)  and  (2); 

.••/=o, 

and  CD  is  constant. 

Hence  (4)  cos^  +  (5)  sin^, 

0  =  i?^  s\n0  -  R^  cos^, 
and  (1)  sin^  —  (2)  cos^, 

0  =  R\  sin^  -  R\  COS0. 

Hence  there  is  no  action  perpendicular  to  the  plane  tln-oiigh 
the  radius  vector  and  the  axis  of  the  cylinder.  Let  R  and  R' 
be  the  horizontal  pressures  in  that  plane  on  the  base  and  at 

U 


290  SOLUTIONS   OF   SBNATE-iroUSE   PROIU.EMS.  [1848. 

the    other   point   of  contact.       Then     (1)  cos^  +  (2)  sin^    and 
(5)  co8^  —  (4)  sin^  give 

11 +  R  =  -  Mrc^' (7), 

and   R  +  E  =  mKvcS' (8): 

a' 

(3),  (7),  and  (8)  are  the  only  equations  for  determining  i?,  R\B^, 
and  B\. 

Tlie  indetcmiinatcness  of  the  problem  arises  from  the  cir- 
cmustancc,  that  there  are  more  pressures  on  the  sphere  than 
are  necessary  to  produce  the  motion  required.  Thus,  there  are 
the  two  vertical  forces  B^  and  B'^  to  support  the  weight,  the 
two  radial  forces  B  and  B'  to  curve  the  path  of  the  centre 
of  the  sphere,  and  the  two,  B  and  B\^  to  oppose  the  tendency 
of  the  sphere  to  rotate  about  a  horizontal  axis  perpendicular 
to  the  radius-vector.  Hence  the  above  equations  contain  the 
sums  of  couples  of  these  quantities.  Considerations  of  elasticity, 
which  prevents  all  such  ambiguities  in  nature,  would  remove 
them  from  the  solution  of  the  problem. 

G.  A  uniform  bent  lever,  whose  arms  are  at  right  angles 
to  each  other,  is  capable  of  being  enclosed  in  the  interior  of 
a  smooth  spherical  surface ;  determine  the  position  of  equi- 
librium. 

Find  also  the  time  of  a  small  oscillation  when  the  position 
of  equilibrium  is  slightly  disturbed. 

Since  the  reactions  of  the  sphere  all  pass  through  the  centre, 
it  is  plain  that  the  resultant  force  of  gravity  upon  the  lever  must 
also  pass  through  the  centre  of  the  sphere ;  hence,  its  centre  of 
gravity  must  lie  vertically  under  the  centre  of  the  sphere. 

Let  C  (fig.  101)  be  the  angle  of  the  lever  ACB,  join  AB-, 
bisect  AB,  AC,  BC,  in  0,  D,  and  E,  and  in  ED  take  the 
point  a,  such  that  EG  :  ED  ::  AC :  AC  +  BC:  join  OE, 
OG,  OD:  G  will  be  the  centre  of  gravity  of  the  lever,  and 
OD,  OE,  will  be  pei-pendicular  to  AC,  BC:  also   0  will  be 


1848.]  RTfJID    DYNAMICS.  291 

the  centre  of  the  sphere.  Hence  we  must  have  00  vertical. 
But  EG:  GD'.:CD'.EC::OE:OD;  therefore  OG  bisects 
the  right  angle  0Z>,  and^C,  5C,  are  equally  inclined  to  the 
horizon. 

^Vhen  the  lever  is  slightly  disturbed  in  its  own  (vertical) 
plane  from  its  position  of  equilibrium,  it  will  manifestly  oscillate 
as  if  it  were  attached  to  an  axis  through  0,  and  the  sphere 
removed.  Hence  we  have  to  find  the  radius  (/>■)  of  gyration 
of  the  straight  lines  AC^  BC,  about  an  axis  through  0,  per- 
pendicular to  the  plane  of  BC^  CA. 

Let  A^j,  k^,  be  the  radii  for  AC^  BCj  respectively :  then 
k;'  =  0D'  +  IAD""  =  EC  +  ^AB' 
=  ¥  +  !«•■',     if  ^  C  =  2a,  BC  =  2h. 
Similarly, 

K  =  a'  +  W, 
and    l>^-J<±JK^ll±^^^±4^±Vl^l(a  +  hf. 

Again,  to  find  0G{=  Z),  the  distance  of  the  centre  of  gravity 
of  the  lever  from  0.     We  have 

._  j^^,    &'mOEG  _      a      j-,^sinOED 
smEOG  ^  TTTh         smEOG 

2-ah 

~  a  +  V 

therefore  the  time  of  a  small  oscillation 

_  2^'       {a  +  hf 

7.  A  section  of  the  surtace  of  a  circular  right  cone  (whose 
axis  is  horizontal  and  vertical  angle  60°)  is  formed  by  a  plane 
pei'pendicular  to  the  slant   side,  so   as  to  contain   the  vertex ; 

U2 


292  SOLUTIONS  OF  si:nati:- HOUSE  i'kouuems.         [1848. 

shew  tliat  when  the  surface  so  cut  off  makes  small  oscillations 

21a 
about  the  axis,  the  length  (jf  the  isochronous  pendulum  =  -— 

(whether  the  elliptic  base  be  included  in  the  surface  or  not), 
a  being  the  length  of  the  pei-pcndicular  drawn  from  the  vertex 
upon  the  elliptic  base. 

Let  k  be  the  radius  of  gj^ration  of  the  section  about  the  axis ; 

r  being  the  distance   of  the  element  hS  of  the  surface  from 
the  axis.     Let  hS  be  projected  upon  a  plane  pei-pendicular  to 
the  axis,  and  hS'  be  the  corresponding  elementary  sm-face ; 
.-.  hS'  =  S>Scos30°; 

,  _  sa^v 

=  the  square  of  the  radius  of  gyration  of  the  elliptic 
projection  on  a  plane  perpendicular  to  the  axis. 

Similarly,  the  radius  of  gyration  of  the  elliptic  base  equals 
the  radius  of  gyration  of  its  projection  on  the  same  plane,  which 
is  the  same  as  the  projection  of  the  whole  section.  To  find 
this  radius  we  must  first  find  the  axes  of  the  elliptic  base  BC 
(fig.  102). 

Since  the  vertical  angle  BAC=SO°^  we  have,  if  AC=aj 

and  if  h  equals  the  semi-axis  minor, 

(2^)"''=  CF.BE=a.2a', 

.'.  ¥  =  K. 

We  may  now  also  find  Oo,  the  distance  of  o,  the  centre 
of  the  base  from  0,  the  point  where  the  axis  pierces  the  base. 

We  have 

Oo^oC-OC 

3i  1 


1849.]  RIGID   DYNAMICS.  293 


•*. 

06'  =  jW, 

and 

Oo"  =  lOd'-- 

-  W,2 

or 

Od  =  ia. 

Let  a',  &',  be  the  semi-axes  of  the  ellipse  BDj  which  is  the 
projection  upon  the  plane  BE  of  the  conical  surface,  as  well 
as  of  the  elliptic  base  BCj 

.-.  2a'  =  BD=  CF+  \[BE-  CF) 

=  a  +  \a  =  la^ 
and    a  =  |a, 

and    h'  =  semi-axis  minor  of  BG 
1 

Hence  the  radius'^  of  g)T*ation  of  BD  about  Oo' 

and  the  length  of  the  simple  pendulum  of  BD  about  the  axis 

17      «■''        ^  , 
=  4X6  W"-^' 


=  T5«  +  i« 

—  2I/T 

—  TB<^* 

Smce  the  length  of  the  simple  pendulum  for  the  conical 
surface  is  the  same  as  for  the  elliptic  base,  it  is  the  same  for 
the  conical  sm^face  alone  and  taken  with  the  base. 

1849. 

1.  If  a  miifonn  inextenslble  string,  in  the  form  of  any 
continuous  curve,  be  subjected  to  an  impulsive  tension  at  its 
extremities,  the  tension  at  any  point  will  vary  directly  as  the 
velocity  communicated  to  that  point  in  the  direction  of  the 
radius  of  absolute  curvature,  and  inversely  as  the  curvature. 


294  iSOLUTlONS   OF   SENATE-HOUSE    I'UOBLEMS.  [1849. 

Let  T  be  the  tension  at  any  point,  then  T  -r-  ^  "^  ji  ^  y  > 

are  the  tensions  in  dircetions  of  the  axes ;  and  since  the  tensions 
arc  impulsive,  wc  have 

difference  of  tensions  at  the  extremities  of  any  small  arc 
QC  velocity  communicated  to  the  arc ; 

or    -^  al  +  I  —r«  as  ec  ^-  (1). 

as  as  at  ^  ' 

Similarly,  |ir+r§*cc| (2), 

%'^-^'i'^-% (^)- 

dx   d'^x      dy   d^y      dz   d'^z 
rp  ^     It    ~d?      'dt    ~dl      It   Is' 

"^    urx\'    Td^    /^v  * 


d'^x  dx      d^y  dy      d^z  dz 

smce  -jT   1-+-?^  T^  +  TiF  -7-=0. 

ds     ds       ds    ds       ds    ds 

Now,  the  direction-cosines  of  the  radius  of  cui'vature  are 
d'^x  d'^y      d'^z 
d£^ ds        ds^ 

d''x\'      (d'yV      fd''zy]  i ' 


Also,  if  p  he  the  radius  of  cui'vature, 

1       (fd'xy       (dW       fd' 


Ai 


p~  Wds'J    "^  [ds'J    "^  Us'J  ^   ' 
,'.  T  Gc  velocity  communicated  to  [xyz]  in  the  direction  of  the 
radius  of  absolute  curvature,  and  inversely  as  the  curvatm*e. 

2.  The  nut  of  a  screw  rests  upon  a  smooth  horizontal  plane, 
over  a  hole  cut  so  as  to  allow  a  free  passage  for  the  screw, 
and  the  screw  descends  through  the  nut  by  its  own  weight: 
detcnnine  the  motion. 


18-19.]  KTGID    DYNAMICS.  295 

At  time  t  let  P  be  the  whole  action  between  the  screw  and 
nut  perpendicular  to  the  thread  of  the  screw,  which  makes  an 
angle  a  suppose  with  the  horizon.     Then 

the  whole  vertical  force  on  the  screw  =  Mg—Fcosa^ 

moment  of  the  whole  horizontal  force =  Pa  sina, 

on  the  nut     =  —  Fa  sina. 

Hence,  if  y  =  depth  of  any  point  of  the  screw  below  a  fixed 
plane, 
ft),  &>',  the  angular  velocities  of  the  screw  and  nut, 

d'^j/  Pcosa 

df 

7  2  ^Ca 

dt 

rrid(o' 

^     dt    ~  M' 

The  geometrical  condition  is,  that  each  two  cori'esponding 
points  of  the  screw  and  nut  in  contact  have  the  same  motion 
perpendicular  to  the  thread; 

dy  ,   . 

.*.  rto)  sma 7"  cosa  =  aco  sma. 

at 

Differentiating  this  equation  and  substituting  from  the  above. 


u 

M 

Pa  sina 

M 

1 

Pa 

sina 

doi  d'^y  _       .       di 


(O 


a  sma  -j cosa  -^  =  a  sma  —r- : 

dt  dt'  dt 

PiC  sin'^'a      P  cosV  Pd^  sin'^a 

whence  P  is  constant,  and  its  value  known :  by  substitution  of 
this  value  we  determine  the  three  required  parts  of  the  motion, 
which  thus  appear  to  be  uniformly  jWcelerated. 

3.  The  centre  of  a  rough  sphere  is  fixed ;  if  another  sphere 
be  placed  on  the  top  of  it  and  just  displaced,  determine  the 
motion  of  both  spheres. 

Let  0,  o,  (fig.  103)  be  the  centres  of  the  spheres  at  the  time  t ; 
0(7,  oc,  the  two  radii  which  in  the  beginning  of  motion  were 


296  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

vertical,  so  that  C,  c,  coincided :  then,  calling  the  different  parts 
and  angles,  as  m  the  figure,  we  have  for  the  equations  of  motion, 

for  the  lower  sphere, 

MU'^=Fa (1). 

for  the  upper  one, 

M'^  =  B  smcf>  -  FcoH<f> (2), 

M'^  =  B  cos<^  +  i^sinc^  -  M[g  (3), 

M'k"^^  =  Fb (4);- 

and  for  the  geometrical  condition  we  must  express  the  circum- 
stance that  the  spheres  roll  without  sliding ; 

...  a{(})-e)  =  b{e'-<f>)  (5). 

Also  we  have 

X  =  [a  +  b)  sin^, 

7/  =  [a  +  b)  coscf). 
Takmg  (1)  ^  —  (4)  «,  we  have 

MFb~-MTa'^^0- 
dt'  df  ' 

whence  we  find 

.,      M¥b  . 

^  M'V'a  ^        suppose. 

Hence  (5)  becomes 

a{(l>-0)  =  nbO  -  b<^y 

a       ^  +  ^  J. 
or     a  = 7  d> ; 

and    &  =  n r  d>. 

a  +  no 

Now  the  expression  for  the  vis  viva  gives  us 

«.(|)V>ri(.+.)'(f)Vr(f)]  =  23/-,,(.+.-,); 


1849.]  RIGID   DYNAMIC'S.  297 

and  substituting  the  above  values  of  d  and  6\  we  get 

=  2M'g{a-\-h-y) 
=  2M'g[a  +  h)  (l-cos(/)) 
which  may  be  written  shortly 


dt 


\m^  (1  -  cos^)  =  7>i'''sin"^^<^ 


dt  1 


f/<^      sin^0 ' 
and    mt  +(7  =  2  log  tan  \^. 

The  constant  C  may  be  determined  by  supposing  ^  to  have 
a  very  small  value  a,  when  i  =  0 ;  whence 

^ ,      tani(f> 

mt  =  2  log  \-^ , 

^  tanja  ' 

and   tan^^  =  tan^a  £*"", 

which  determines  </»,  and  thence  6  and  ^',  in  terms  of  t. 

4.  What  must  be  the  angular  velocity  of  a  horizontal 
cylinder,  in  order  that  a  heavy  string  of  given  length  attached 
to  it  may  be  just  wound  up  ? 

Let  I  be  the  whole  length  of  the  string,  x  that  of  the  part 
hanging  doAvn,  /u,  the  mass  of  a  imit  of  length ;  T  the  tension 
of  the  rope  at  the  point  of  its  contact  with  the  cylinder.  Then 
for  the  equation  of  motion  of  the  cylinder,  and  that  part  of 
the  rope  coiled  on  it, 

{Mk^  +  fi{l-x)d^}'^  =  -Ta (1); 

and  considering  the  part  of  the  string  hanging  down  as  one 
mass,  the  coordmate  of  a  fixed  point  of  which  (the  extremity)  is  Xj 

and     1  —  X  =  (10. 


298  Solutions  of  senate-iioiise  problems.         [1841). 

Now  (1)  —  (2)  a  gives 

[Mk^  +  /i  (/  -  a?)  a'}  -^-  fiax-^,  =-  figax, 

(Px  fPx 

J/A;"  +  fi[l  —  x)a\  -T^  +  /^a  x  -j^  =  figa  x, 

d'^x 
or    {MP  +  fild^)  -jj  =  f^ai'x. 

dx 
Multiply  by  2  -^ ,  and  integrate, 

.-.  [Mk'  +  fild')  ('^Y  =  fjigd'x'  +  C: 

doc 

and  ;7-  =  ^)  when  a;  =  0 ;  .'.  (7=0; 


.-•.  [Mh' ^  fild')  {^\   =figd'x\ 


Hence,  in  the  beginning  of  motion,  when  £c  =  /, 

dd__\^dx_        {H',g)'  I 
dt~~  li~dt~  [MW  +  iiM)^  ' 

which  is  the  required  angular  velocity. 

5.  A  heavy  rod  is  suspended  from  a  fixed  point  by  two 
inextensible  strings  without  weight,  the  strings  and  the  rod 
forming  an  equilateral  triangle ;  if  either  of  the  strmgs  be  cut, 
dotemiiue  the  initial  tension  of  the  other. 

Let  the  figure  (104)  represent  the  position  of  the  beam  at 
the  time  t  after  the  string  has  been  cut ;  GN  being  the  vertical 
line  through  the  point  of  support.  Hence  the  equations  of 
motion  will  be 

.lf^^=rsin^ (1), 

Jf  ^  =  Mg  -  TcosO (2), 

Mk^^  =  -  Ta  sin(6'-f  <^) (3). 

Also  the  geometiy  gives  us 

X  =  2rt  sin^  —  a  sin^, 
y  =  2a  cos^  +  a  cos(f). 


1849.]  RIGID   DYNAMICS.  299 

Hence,  differentiating  twice,  we  find 

d'x   ,   ^      d'y       ^  ^    (ddV  .   ,„      ,.d"(i) 

-aco8(6>  +  (^)(^'^ 

T 
=  g  cos^  —  -^cos2^, 

by  (1)  and  (2). 
Hence  we  find,  by  substituting  the  value  of  -j^  from  (3), 

-^{co82^+|j8in"''(^+</))}=5rcos^  +  2a(-^)  +acos(^+</))f-^)   . 

Now,  in  the  beginning  of  motion,  — -  =  0,  -y^  =  0,  ^  =  SO"", 
and  0  =  90°:  let  7^,  be  the  initial  tension ; 

34 


and  F  =  ^a^ ; 


the  required  tension. 


„  /I       9\       „  34 
or     T^'^-^Mg, 


6.  A  man  standing  in  a  swing  is  set  In  motion :  shew  that 
he  can  accelerate  the  motion  and  increase  the  arc  of  oscillation 
by  crouching  and  rising  In  the  swing ;  and  prove  that  the  effect 
will  be  greatest  If  he  crouch  when  the  swing  Is  at  the  highest 
point,  and  rise  when  It  Is  at  the  lowest  point  of  its  arc  of 
oscillation. 

Since  the  ropes  of  the  swing  are  not  supposed  to  slacken 
or  bend,  we  may  suppose  them  to  become  rigid,  and  rigidly 
connected  with  the  swing. 

If  the  man  do  not  crouch  and  rise,  the  arc  of  oscillation  will 
be  unaltered,  the  effect  of  gravity  being  to  accelerate  the  motion 
while   he   Is  descending,    and    to    retard    It    while    he    ascends. 


300  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

Now,  if  the  man  rises  when  the  swing  is  at  its  lowest  point, 
the  moment  of  the  force  of  gravity  on  him  about  the  axis 
through  the  points  of  support  of  the  swing  is  diminished,  and 
the  motion  less  retarded  tlian  it  would  have  been  if  he  had 
retained  a  mean  position ;  hence  the  swing  will  rise  higher  than 
it  otherwise  would :  if  he  crouches  when  at  the  highest  point 
of  the  arc  of  oscillation,  the  motion  will  be  more  accelerated 
while  the  swing  descends  than  it  would  have  been  if  he  had 
remained  in  a  mean  position ;  hence  the  velocity  at  the  lowest 
point  will  be  increased  on  account  of  his  having  both  crouched 
and  risen ;  hence  the  arc  of  oscillation  will  be  mcreased  by 
such  a  motion  of  his  body. 

It  is  evident  that  it  will  be  most  increased  if  he  rises  at 
the  lowest  and  crouches  at  the  highest  point  of  the  arc  of 
oscillation. 

In  addition  to  the  above  reasons  w^hy  the  supposed  motion 
of  crouching  and  rising  will  increase  the  arc  of  oscillation,  is 
another,  viz.  that  the  principle  of  the  conservation  of  areas 
must  hold  during  the  sudden  motion  of  rising  at  the  lowest 
point.  For  during  that  motion  both  the  forces  on  the  man, 
viz.  gravity  and  the  upward  pressure  of  the  swmg,  may  be 
considered  as  acting  in  a  vertical  direction,  that  is,  normally 
to  his  instantaneous  direction  of  motion.  The  consequence  will 
be,  that  his  linear  velocity  will  be  increased  as  he  rises,  and 
therefore  approaches  the  horizontal  axis  through  the  points  of 
support  of  the  ropes. 

If  the  swing  be  supposed  to  have  mass  this  effect  will  be 
diminished,  since  his  rising  will  not  so  much  raise  the  common 
centre  of  gravity  of  himself  and  the  swing.  This  diminution 
of  the  effects  of  the  principle  of  conservation  of  areas  will  be 
practically  caused  by  a  change  of  the  friction  between  the  swing 
and  his  feet,  which  will  for  the  mstant  retard  his  motion  more 
than  it  usually  does. 

The  above  reasoning  has,  of  course,  no  place  as  applied  to 
his  crouching  when  at  the  highest  point  of  the  arc  of  oscillation, 
since  he  is  then  describing  no  areas  at  all  about  the  horizontal 
axis  through  the  points  of  support. 


1849.]  RIGID   DYNAMICS.  301 

7.  A  circular  hoop  rests  upon  a  smooth  horizontal  plane 
with  a  particle  at  its  lowest  point,  and  receives  a  horizontal 
velocity  of  projection  V  in  its  own  plane :  find  the  value  of  V 
in  order  that  the  particle  may  just  rise  to  the  height  of  the 
centre  of  the  hoop. 

DcteiTnine  the  motion  when  V  is  greater,  and  also  when  it 
is  less  than  this  value ;  and  find  the  time  of  an  oscillation  of 
the  particle  in  the  hoop  when  V  is  small. 

Let  P  be  the  position  of  the  particle  in  the  hoop  (fig.  105) 
at  the  time  t  from  beginning  of  motion.  Let  AN=  ic,  NP=y^ 
be  the  coordinates  of  P  referred  to  A^  its  position  at  projection, 
as  origin,  AM  =  x.  The  principle  of  the  conservation  of  the 
motion  of  the  centre  of  gravity  in  a  horizontal  direction  gives  us 

M^  +  M'^  =  M'V (1). 

The  expression  for  vis  viva  is 

Also  x  —  X  =  a  sin0, 

y  =^  a[l  —  cosO) ; 

dx       dx  rt  ^^  /.,\ 

.-.   -: =-  =  a  cosy  -r  (3), 

dt       dt  dt  ^  ^' 

dy  .  add  . 

-f  =  a  smO  -y- (4). 

dt  dt  ^  ' 


-Y-  =  a  s'mO  -T- 

dt  dt 

From  (1)  and  (3), 

(31^ M')  §  =  M'[V- a  cose '-^;), 

{M+M')  ~  =  M'V+  MacosB^. 
ince  equation  (2)  becomes 

IM^'  H'  ( ''-  "  -^^  I)'  +  (^'  '^+  ^o  •=-»  f )} 


302  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [18-49. 

■ja 
If  the  particle  just  rises  to  the  height  of  the  centre,  -^  =  0, 

when  0  =  00°  and  y  =  a\ 

-  r=(^-^/(2,.).. 

AVhen  V  is  greater  than  this  vahie,  equation  (5)  gives  the 

fid 
height  to  which  it  will  rise  before  -r-  =  0,  viz. 

y  ~  M+  M' '  2(7  ■ 

At  this  time  the  particle  is  moving  horizontally  with  the 
same  velocity  as  the  hoop :  it  will  now  fall  down  in  a  parabolic 
path  and  will  strike  the  hoop  at  a  point  at  the  same  distance 
below  the  horizontal  diameter  as  the  point  at  which  it  left  the 
hoop  is  above  it. 

If  V  is  not  sufficiently  great  to  make  it  rise  to  the  height 
of  the  centre,  it  will  rise  to  the  height 

M'        V^ 

and   since  the   above   equations   apply   for   both   directions   of 

motion  of  the  particle,  we  see  that  it  will  continue  to  oscillate, 

rising  on  both  sides  of  the  vertical  diameter  to  the  above  height. 

(19 
If  V  be  very  small,  Q  and  -j-  will  be  very  small :   in  this 

case  differentiate  (5) ; 

-P — ^f^  ]  cos''0.2  -^  -r-2  smO  cosB    -^     i  =  -  2q  -^ 
\I+M   [  dt   dt  \dfj  )  •'  dt 


'  M+M' 


or 


M'a      \,  d'^d         fddV) 

ifTM'f-^^W-^[dt.)\='-^^^ 


=  —  2ga  smt;  -j-  , 


3f  + 

dO 


or  omitting  OP  and  9  ^  , 


1849.]  Kin  1 1)   DYNAMICS.  303 

df  ^       M'        a'^         ' 
the  equation  of  oscillator}^  motion,  the  time  of  oscillation  being 

/     M'        a\i 

8.  A  heavy  lamina,  in  the  form  of  an  equilateral  triangle, 
suspended  fi'om  a  fixed  point  by  three  equal  strings,  is  drawn  a 
little  aside  from  its  horizontal  position  (the  strings  being  all 
stretched) ;  its  centre  of  gravity  then  receives  a  small  honzontal 
velocity  of  projection  perpendicular  to  the  plane  in  which  the  dis- 
placement was  made,  while  at  the  same  time  a  velocity  of  rotation 
is  connnunicated  to  it  in  its  owti  plane :  detemiine  the  motion. 

The  motion  of  rotation  of  the  lamina  in  its  own  plane  is 
evidently  independent  of  the  motion  of  its  centre  of  gravity,  and 
will  continue  uniform. 

By  the  principle  of  the  superposition  of  small  motions,  the 
oscillations  of  the  centre  of  gravity  in  the  two  perpendicular 
planes  will  be  independent  of  each  other. 

The  equations  of  these  small  motions  will  be 
6  =  6^  cosnt 
and     (f)  =  ^^  sinnf^ 
6^  and  ^,  being  the  semi-arcs  of  oscillation ; 

and  the  centre  of  gravity  will  move  veiy  nearly  in  a  small 
ellipse  about  its  position  of  equilibrium  as  centre,  with  axes 
rB^  and  r(^^,  r  being  the  distance  of  the  centre  of  gravity  from 
the  point  of  suspension. 

9.  A  man  hangs  by  a  rod  which  swings  in  a  vertical  plane : 
compare  the  exertion  required  to  raise  him  from  one  given  point 
of  the  rod  to  another,  1st,  when  he  draws  himself  through  a 
given  small  space  always  when  the  rod  is  vertical,  and  2ndly, 
if  he  makes  the  effort  when  the  rod  is  at  its  greatest  inclination 
to  the  vertical. 

See  Prob.  6,  1849. 


304  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1849. 

Let  V  be  the  velocity  of  tlic  man  at  the  lowest  point  before 
he  begins  to  raise  himself,  2a  the  arc  of  oscillation:  then,  if 
a  be  his  original  distance  from  the  point  of  support, 

v^  =  rt  (1  —  cosa). 

Now,  if  he  raises  himself  always  through  the  given  small 
space  B  at  the  lowest  point  of  the  arc,  the  velocity  at  the  lowest 
point  will  receive  a  sudden  increase;  and  it  will  be  with  this 
increased  velocity  that  he  will  next  airive  at  the  lowest  point: 
the  arc  of  oscillation  will  also  continually  increase. 

Let  v^  be  the  velocity  with  which  the  man  arrives  for  the 
r**»  time  at  the  lowest  point  of  the  arc,  when  he  raises  himself 
through  the  r^^  small  space  8.     The  exertion  of  doing  this 


i        a-{r-l)8j 


To  detennine  v^.  The  relation  between  v^  and  v,._^  is  given 
us  by  the  equation 

v^{a-{r-l)S]=v,._^{a-{r-2)B], 

which  expresses  the  conservation  of  areas  during  the  man's  rise 

through  the  small  space  S.     We  may  hence  deduce  the  equation 

v^  [a—  (r—1)  B]  =  va. 

This  equation  we  may  also  derive  from  the  consideration, 
that  the  man  returns  each  successive  time  to  the  point  when 
he  raises  himself  with  the  velocity  with  which  he  quitted  it; 
and  therefore  we  may  consider  that  he  raises  himself  by  one 
effort  through  the  space  (r— 1)S,  the  conservation  of  areas 
holding  all  the  while.  This  consideration  gives  us  the  above 
equation  immediately.     We  thus  have 


a-{r-l)8       {a-{r-l)BY' 
Hence  the  r"^  exertion 

Let  us  call  (r— 1)S,  x',  then  we  may  call  S,  cfe,  and  we 
shall  have,  as  an  approximation  to  the  true  result,  supposing 
B  extremely  small, 


1849.]  RIGID  DYNAMICS.  305 

whole  exertion 

=  M\gh  +  \vV[j^,^^-l:) 

If  the  man  raises  himself  at  the  highest  point  of  the  arc, 
the  arc  of  oscillation  will  remain  mialtered,  but  the  velocity 
at  the  lowest  point  will  be  increased  after  each  effort.  Hence 
every  exertion  will  be  the  same,  viz.  Mg  sin  a.  8,  and  the  whole 
exertion  Mg  sina.^- 

Hence  the  ratio  of  the  two  exertions 

/         v^[2a-h)\ 

=  I'  +  ^FF^v '"'''°'- 

10.  A  semicircular  board,  moving  in  its  own  plane  without 
rotation,  and  with  Its  cm'ved  boundaiy  foremost,  comes  in  con- 
tact with  a  smooth  fixed  obstacle :  determine  at  what  point  the 
impact  should  take  place  in  order  that  the  angular  velocity 
generated  may  be  the  greatest  possible. 

Let  P  (fig.  106)  be  the  point  where  the  Impact  should  take 
place,  the  radius  CP  making  an  angle  6  with  CD  the  bisecting 
radius.  Let  P^be  the  velocity  (in  CD)  of  the  centre  of  gravity 
of  the  board  before  impact,  t" ,  v\  those  parallel  to  CD  and  CB 
after  Impact,  ^  the  angular  velocity  after  Impact,  R  the  Im- 
pulse :  then,  if  G  be  the  centre  of  gravity  and  C6^  =  a, 

V  =  V  —  ■zTi.coaO. 
M        ' 

^'  =  ^«in^>  ) (A), 

_  Ra  %md 


306  SOLUTIONS   OF   SENATE-HOUSE   PIIOBLEMS.  [1849. 

Also  we  have  the  geometrical  condition,  that  P  must  have 
no  motion  in  the  direction  CP  after  impact ;  whence 

V  co3^  -  v  sin^  -  r^PO  amGPC  =  0, 
or   V  cos^  —  v  sin6  —        wa  sin^        =  0. 

Finding  r,  v',  from  equations  (A)  in  terms  of  w,  and  substi- 
tuting in  this  last  equations,  we  get 

fV-  —  cot^]  cos^  -  —  sin^  -  ^'x  sln^  =  0, 
or     V  cos^  —  ( —   ^— ^  +  a  smO  )  •cj  =  0 


.'.     CT   = 

V  sinO  COS0 

h  a  sm  y 

a 

ich 

I  is 

to  be  a 

maximum  by  the  variation  of  0. 

N( 

DW, 

a 

4 
=  37r' 

a, 

k' 

=  W 

-«'^; 

.'.    in 

Va 

sin0  COS0 
-  a'  cos'^0 

Va  sin  2^ 

«'  -  a'  (1  +  COS  20) 
Taking  the  logarithmic  differential, 

a'  sin  20 


0  =  cot20  - 


«'-a'(l  +  cos20)' 
or    (a'  -  a')  cot 20  -  a''  cosec20  =  0, 

or    cos20  = 


a   —  a 
16 


97r'  -  16  ' 
which  determines  the  position  of  the  point  P. 

11.    A   imiform   solid   cylinder   is   revolving  with  a  given 
angular  velocity  about  its  centre  of  gravity,  which  is  fixed ;  the 


1841).]  KKilD    DYNAMICS.  307 

cylinder  then  receives  a  blow  of  given  intensity  in  a  direction 
perpendicular  to  the  plane  in  which  the  axis  moves:  determine 
the  subsequent  motion. 

Since  any  section  of  the  cylinder  through  its  axis  is  a  prin- 
cipal section,  the  blow  takes  place  in  a  principal  plane,  and 
therefore  only  generates  a  velocity  about  the  axis  (that  of 
X  suppose)  perpendicular  to  the  axis  of  previous  rotation  (that 
of  j/),  and  the  axis  of  the  cylinder  (that  of  2). 

Hence,  if  A  be  the  moment  of  inertia  about  the  axes  of 
X  and  y^  C  that  about  the  axis  of  z ;  and  w,,  eo^,  Wg,  be  the 
angular  velocities  about  these  respective  axes  at  any  time  t 
after  impact,  we  have  as  equations  of  motion, 

A'^-{C-A)a>.^^o.^  =  0, 

C^-{A-A)co^<o^  =  0, 

or    ^^  =  0, 
df  ' 

and    0)3  =  constant 

=  0,  since  that  is  its  original  value. 

Hence  also        tUj  =  constant 

=  that  generated  by  the  blow 

=  — 7- ,  if  i?  be  the  blow,  a  the  distance  from 

A  '  ' 

the  centre  of  its  point  of  application, 
and    (o,,  =  constant 

=  its  original  value  before  impact. 

Hence  the  cylinder  revolves  unifonnly,  and  the  instantaneous 
axis  is  fixed  in  it,  viz.  in  the  plane  xy ;  hence  this  axis  is  also 
fixed  in  space,  and  the  axis  of  the  cylinder,  as  before,  sweeps 
out  a  plane. 

x2 


308  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

12.  A  Rolld  cone  is  suspended  by  its  vertex  from  a  point 
in  a  perfectly  rough  wall :  if  the  cone  be  slightly  displaced  from 
its  equilibrium  position,  the  surface  remaining  in  contact  with 
the  wall,  determine  the  time  of  a  small  oscillation. 

Let  0  be  the  angle  which  the  line  of  contact  of  tlie  wall 
and  cone  make  with  the  vertical  at  the  time  t]  to  the  angular 
velocity  at  the  same  time  about  the  line  of  contact ;  0^  the 
original  value  of  ^;  h  the  height  and  2a  the  vertical  angle 
of  the  cone :  then,  since  the  line  of  contact  is  the  instantaneous 
axis,  the  equation  of  vis  viva  gives  us 

Jf/c'V  =  2Mg.^h  cos  a  (cos0  -  cos^J, 

2      3    qh  cosa  ,       ^  n\ 

or  <"   =  n         7,>ii —  (cost/—  cos^J. 

Now,  to  connect  w  and  0,  we  have  two  expressions  for  the 

rlB 
motion  of  the  centre  of  the  base,  viz.  h  sina.w  and  h  cosa .  -r- ; 

dO      ^ 
.-.  0)  =  ^  cota; 

fd()\^      3    qh  sin^a  ,      ^  n\ 

•■•  [it)  "2  'ks^f'^''^^ -"><); 

.*.    -jY  =  -  J       \n      sin0  ,  I  the  length  of  the  side ; 

or  if  the  motion  be  very  small, 

d''0      3    «7/sin'''a  „      ^ 

k' 
Therefore  the  time  of  a  small  oscillation  =  47r 


(3^Z)^  sina 


13.  An  indefinitely  great  number  of  indefinitely  thin  cylin- 
drical shells,  just  fitting  one  within  another,  are  revolving  with 
different  angular  velocities,  but  in  the  same  sense,  about  their 
common  axis;  also  the  angular  velocity  of  each  shell  is  pro- 
portional to  a  positive  power  (the  n"^)  of  its  radius,  and  that 
of  the  outermost  shell  is  tu.     Prove  that  if  the  system  of  shells 


1849.]  RIGID   DYNAMICS.  309 

be  suddenly  united  into  a  solid  cylinder,  the  cylinder  will  revolve 

about  its  axis  with  the  angular  velocity . 

If  the  bodies  composing  the  Solar  system  were  suddenly  to 
become  rigidly  connected,  explain  what  the  nature  of  the  sub- 
sequent motion  would  be. 

If  0)^  be  the  angular  velocity  about  the  axis  of  any  particle,  at 
a  distance  r  from  the  axis,  the  area  described  by  it  in  the  time  t 

n  n+!j 

7'  r 

a  a 

if  a  be  the  radius  of  the  outer  shell. 

Hence  the  sum  of  the  areas  described  by  all  the  particles  in 
the  same  cylindrical  shell,  of  thickness  Sr, 

=  TTO)  -%-  t .  Br. 
a 

Hence  the  sum  of  all  areas  described  by  all  particles  in  time  t 


t   r 
—      ' 

«    Jo 


a 
=  rrmt 


«  +  4 


If  w  be  the  angular  velocity  after  uniting,  the  sum  of  the 
areas  described  by  all  the  areas  in  an  equal  time 

=  TT'ot  I  r^hr 


•'  0 


=  TTWC  .  — 
4 


By  the  principle  of  the  conservation  of  areas,  the  two  above 
sums  of  areas  must  be  equal,  or 


a'  or 

TT-art  —   =  TTtOt 1 

4  n  +  4  ' 


4(i) 


«  +  4 


310  soLLTioNS  OF  sENATK-iiuL'sE  rKOBLi::>:s.  [1850. 

In  the  Solar  system  the  areas  of  the  orbits  vary  as  the  square 
of  the  mean  distances,  and  the  periodic  times  squared  as  the  mean 
distances  cubed,  and  therefore  the  mean  angular  velocities  vary 
as  the  square  roots  of  the  mean  distances.  Hence,  if  all  the 
bodies  of  the  system  were  suddenly  to  become  rigidly  connected, 
the  bodies  nearer  the  sun  would  have  their  motions  suddenly 
accelerated,  and  those  furthest  from  it  would  be  suddenly  re- 
tarded ;  after  which  all  would  proceed  with  a  common  unifonn 
and,  so  to  speak,  an  average  angular  motion. 

1850. 

1.  A  parallelogram,  whose  centre  is  fixed,  is  rotating  about 
one  of  its  principal  axes  in  its  plane ;  find  how  it  must  be  stiiick 
that,  after  the  blow,  it  may  rotate  with  the  same  angular  ve- 
locity about  the  other. 

Since  the  effect  of  a  blow  upon  the  velocity  of  rotation  of 
the  body  about  a  principal- axis  depends  only  upon  the  moment 
of  the  blow  about  that  axis,  it  is  plain  that  the  blow  must  in 
this  case  be  perpendicular  to  the  plane  of  the  parallelogram, 
and  Its  moments  about  the  two  principal  axes  in  its  plane  must 
be  equal  and  be  due  to  the  velocity  of  rotation  already  existing 
about  one  of  them,  and  in  a  direction  to  destroy  it. 

Let  A,  B^  be  the  moments  of  inertia  about  these  principal 
axes,  0)  the  velocity  of  rotation  about  A  before  the  blow  (/) 
IS  given,  x,  «/,  the  coordinates  of  the  point  of  application  of  the 
impulse ; 

Am 


y 


Bm 


which  equations  determine  the  point  of  application  of  the  im- 
pulse. 

2.  Three  equal  smooth  spheres  (radius  r)  are  placed  together 
on  a  horizontal  plane,  and  kept  in  contact  by  a  string  passed 
round  them  in  the  plane  of  their  centres.  A  cone  of  given 
weight  [W]   and  vertical  angle  (2a),   is  placed  between  them 


1850.]  nroiD  dynamics.  311 

80  that  its  axis  is  vertical :  find  the  tension  of  the  string ;  and 
if  the  string  be  suddenly  cut,  find  when  the  cone  will  strike 
the  plane. 

If  R  be  the  pressure  between  any  sphere  and  the  cone,  we 
have  for  the  equililibrimn  of  the  cone 

3i2sma  =  TF; 
and  for  that  of  any  sphere, 

i?cosa  =  2  TcosItt 

„      FTcota 

the  required  tension. 

At  the  time  t  after  the  string  is  cut,  let  y  be  the  height  of 
the  vertex  of  the  cone  above  the  plane,  x  the  distance  of  any 
sphere  fi'om  the  axis  of  the  cone :  the  equation  of  vis  viva  gives 
us,  if  W  be  the  weight  of  any  sphere, 

where  y^  equals  the  height  of  the  vertex  of  the  cone  in  the 
position  of  equilibrium. 

We  have  also  to  express  the  condition,  that  the  motion  of 
the  point  of  any  sphere  in  contact  with  the  cone  in  the  direc- 
tion perpendicular  to  the  generating  line  of  the  cone  through 
the  point  of  its  contact  with  that  sphere,  is  equal  to  the  motion 
of  that  point  in  the  same  direction :  or 

dx  dii    . 

T-  cosa  =  — J-  sma : 
dt  dt  ' 

dx  dy  ^ 

.'.   -J-  = r  tana. 

dt  dt 

Hence  the  above  equation  becomes 

( W+  3  W  tan'^a)  (^)'  =  2  Wg  {y,  -  y) : 
or  differentiating, 


^ Wg_ 


de  ~       Tr+3ir'tan''a 


=  —  /'  suppose, 


312  SOLUTIONS   OF   SENATE-HOUSE    FKOliLEMS.  [1850. 

a  constant   retarding  force.     And  the  cone   starts   from  rest; 
therefore  the  time  of  describing  the  space  y^,, 

To  find  y^.  From  fig.  107  it  is  evident  that  OiV,  the  per- 
pendicular on  the  axis  of  the  cone  from  0,  the  centre  of  any 
sphere  in  the  position  of  equilibrium,  is  the  distance  of  any 
angular  point  of  any  equilateral  triangle  of  side  It  from  its 
centre  ; 

/--»  AT  TT  2 

.-.   6>A  =  r  sec  -  =  p  r ; 
.*.   OT  =  ON  seca  =  ^  ?'  seca, 
and   PT  =  ( -5  seca  -  1 J  r ; 


CT  =  PT  coseca 
2 


-J  seca  coseca  —  coseca  j  r ; 
.-.  CN=^  CT-NT=CT- OTsma 

I  r^  (seca  coseca  -  tana)  —  coseca^  r 

=  f-i  cot  a  — coseca]  r, 
and    y^  =  r  —  CN 

=  f  1  -f  coseea  —  —^  cota  j  r-; 

1(1+  coseca  -  -j  cota)  ( W-\-  3  W  tan'a)  r  j 

3.  Two  equal  particles  of  mass  m  are  fixed  at  the  ex- 
tremities of  the  axis  of  a  prolate  spheroid,  of  which  the  mass 
is  il/,  the  eccentricity  of  the  generating  ellipse  being  e.  The 
spheroid  is  struck  by  a  couple  and  then  left  to  move  freely; 


1850.]  RIGID   DYNAMICS.  313 

shew  that  throughout  the  motion  it  will  constantly  have  contact 
with  a  single  plane,  if 

m  =  ■^o^e^ 

The  spheroid  will  manifestly  have  contact  with  a  fixed  plane 
parallel  to  the  invariable  plane,  if  it  be  similar  to  the  momenta! 
spheroid  of  the  system  consisting  of  the  spheroid  with  the  two 
masses  [m)  at  its  poles.  Let  -4,  jB,  be  the  moments  of  inertia 
of  this  system  about  its  axis  of  figure  and  an  axis  through  Ita 
centre  perpendicular  to  its  axis  of  figure.  Hence,  if  a,  5,  be  the 
semiaxes  of  the  generating  ellipse,  the  condition  of  contact  with 
a  single  plane  is 

Ad'=^Bh' (1). 

Now  A  =  lMb\    B==\M{a^  +  h')-\-2ma\ 

and  condition  (1)  becomes 

or     2ma'  =  lM{a'-b')- 

=  me\ 

4.  A  small  arc  of  a  hoop  is  removed  and  replaced  by  two 
small  straight  lines,  tangents  to  the  circle  at  the  ends  of  the  arc, 
their  mass  being  so  disposed  that  the  centre  of  gravity  remains 
still  at  the  centre  of  the  hoop.  If  the  hoop  be  now  rolled  along 
a  horizontal  plane,  sufficiently  rough  to  prevent  sliding,  with  an 
angular  velocity  a>  not  great  enough  to  make  it  leap,  shew  that 
motion  will  never  cease  unless 

.       f  o)''*       d^  +  F  cos'^a 

\2ga  '  (1  —  cosg)  cos  a 


be  a  whole  number ;  where  a  is  the  radius  of  the  hoop,  k  its 
radius  of  gyration,  and  2a  the  angle  subtended  by  the  arc 
removed. 


314  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

The  motion  of  rolling  on  the  circular  rim  will  be  unifoiTn : 
hence  the  angular  velocity  at  the  time  of  impinging  on  the  apex 
of  the  tangent  will  be  to.  To  determine  the  motion  immediately 
after  this  impact.     See  fig.  1 08. 

Let  F^  F'  be  the  impulsive  actions  between  the  apex  and 
the  plane  along  the  plane  and  pei'pendicular  to  it;  v^,  v^  the 
velocities  of  the  centre  of  gravity  in  the  same  directions,  and 
■C7  the  angular  velocity  after  impact.  The  equations  for  finding 
y^,  V  .  and  -nr,  are 

F 

F' 

MFzj  =  Mk^m  +  Fa  -  F'a  tana. 

Also,  to  express  the  condition  that  the  apex  must  be  at  rest 
after  impact,  we  have 

v^  —  aw  =  0, 

Vy  —  a  tana.-cT  =  0. 
Hence  we  have 

Jc^z:  =  Jc^oi  +  a^  (o)  -  w)  —  (a  tana)^.^, 

a'(l+tan''a)  +  k' 

=  nco  suppose. 

;Next,  to  consider  the  continuous  motion  of  turning  about 
the  apex :  let  6  be  the  inclination  to  the  horizon  of  the  radius 
to  the  apex  at  the  interval  t  after  impact.  Then,  taking  the 
equation  of  moments  about  the  apex, 

M{/c^  +  c^  sec'^a)  -yy  =  —  Mga  sec  a  cos^ ; 

'd6\ 


(da\  2qa  sec  a      .    ^       ^ 

\dt  J  F  +  a"  sec^a 


2qa  seca     ,  .    ^  > 

7  2,    2 2-  sm  e  -  cos  a 

k  -f  a  sec  a  ^  ' 


1850.]  RIGID   DYNAMICS.  31,') 

\dtj  k~  +  a'  sec'a  ^  i       \  n 


a     1  A'^^^' 


In  passing  from  a  motion  of  rotation  about  the  axis  to  a 
motion  of  rolling  on  the  circular  rim,  there  is  no  impulsive 
motion ;  hence  the  angular  motion  at  the  time  of  the  second 
impact  of  the  apex  on  the  plane  is  w  or  noa.  Similarly,  the 
angular  velocity  at  the  time  of  the  >•'*'  impact  is  ifca. 

Now,  if  the  hoop  ever  comes  to  rest  it  must  be  by  just 
balancing  on  the  apex  :  suppose  this  happens  when  it  is  rolling 
over  the  apex  for  the  ?«"'  time;  then  equation  (1)  shews  that 
we  must  have 

=  w    ft)  -  ^-^ — ^ —   1  -cosa), 

/c   +  a^  sec'a  ^  '^ 


o     1  if       2(7a  sec  a         ,, 

or     2m  log?i  =  log  ijj^-^, r^-^z    l-cosa 

'^  [[k  +  a  sec  a)  w 


ft)"      («^  -f  P  cos''^a) 


lof 

^  I2qa    (1  -  cosa) cosa 
or     m  =  ^\  -  1 

2log|n-^^^tan^a| 

by  inverting  both  the  quantities  under  the  logarithmic  sign. 
Hence,  if  this  expression  for  m  be  a  whole  number,  the  hoop 
will  come  to  rest  as  it  is  rolling  over  the  apex  for  the  ?»"•  time : 
if  it  be  not  a  whole  number  it  will  never  come  to  rest. 

5.  Two  similar  homogeneous  cords  are  similarly  stretched, 
and  one  of  them  loaded  at  its  middle  point  with  a  small 
weight  /x;  shew  that  the  fundamental  note  of  the  loaded  cord 
will  be  lower  than  that  of  the  other,  and  that  if  t  denote  the 
time  of  vibration  of  the  loaded  cord  for  any  possible  note,  the 
values  of  t  are  given  by  the  equation 

tan  H  fiVl  =  A  Uxl< 


wlicrc  /  is  the  length  of  the  cord,  •  the  mass  of  a  unit  of  length. 


316  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

and  T  its  tension.  Under  what  initial  circumstances  will  both 
strings  sound  the  same  note? 

See  Duhamel  in  the  Journ.  de  VEcole  Polytech.  tom.  xvii. 

It  is  evident  that  when  the  loaded  cord  is  sounding  its  fun- 
damental note,  its  two  halves  meet  in  a  salient  angle  pointing 
from  the  line  joining  its  extremities.  Thus  the  two  halves  are 
portions  of  the  trochoid  which  a  longer  cord  would  assume  in  vi- 
brating, and  will  vibrate  in  the  same  maimer  as  if  they  actually 
were  parts  of  such  a  trochoid :  hence  the  loaded  cord  is  virtually 
longer  than  the  unloaded  one,  and  capable  of  a  deeper  note. 

Let  ic,  3/,  be  the  coordinates  of  any  particle  of  the  string  at 
the  time  f,  referred  to  the  line  joining  the  extremities  of  the 
string,  and  a  line  through  its  middle  point  perpendicular  to  it 
and  in  the  plane  of  vibration,  as  axes  of  x  and  y.  The  equation 
of  transversal  vibrations  will  be  (Poisson,  Mecanique^  n°.  490) 

,    .    rmr  ,. ,       >         mir 
y  =  h  sm  -^—  [\L  —  x]  cos  ^—  at^ 

A  A< 

where  h  is  the  greatest  value  of  y  for  any  particle,  a  =  ( - )  ,  and 

\  a  quantity  to  be  determined  by  the  circumstance  that  the 
middle  point  of  the  string  is  attached  to  the  weight  [i. 

Let  y  be  the  ordinate  of  this  weight  at  the  same  time  t : 
the  equation  of  motion  of  ^l  is 


Now 


tan 


d'y 

= 

KIL 

= 

hrriTT        rmr  I         rmr 
—  2t  .  — —  cos  — -  -  cos  — -  at] 

A,                 A      2             A 

•••  y 

= 

2t    h\      1         rmr  I         rmr 
— .  —  .  —:.  cos  — —  -  cos  -T—  at. 
fi     rmr    a            A    2           A 

y 

= 

Jt=o 

= 

J    .    rmr  I         rmr 
h  sm  — -  -  cos  — -  at : 

A      <S              A 

rmr  I 

2t      \       1 

\    2 

fj>  '  mir  '  d^ 

= 

2c        \                .                ,         T 

—  .  —  ,     since  a  =  -  . 

fx.     ymr                           £ 

1850.J  RIGID    DYNAMICS.  317 

Now,  let  t  represent  the  time  of  vibration ; 
_  2X,    1  _  2\     /£ 

m  '  a       m  '  \T 

and  the  above  equation  becomes,  by  eliminating  \, 

the  required  equation  for  the  determmation  of  all  possible  values 
of  «. 

The  value  of  t  answering  to  the  fundamental  note  is  its 
greatest  value;  it  is  plainly  such  that 

T-l;)  "2- 

Now  suppose  fi  indefinitely  small,  or  the  weight  removed, 
the  value  t'  of  t  then  answering  to  the  fundamental  note  is 

TT?      /S\4  _  TT 

T-[t)    -2' 

and  therefore  t  >  t\  or  the  fundamental  note  is  lower  for  the 
loaded  than  for  the  unloaded  cord,  as  shewn  above. 

The  two  cords  will  evidently  sound  the  same  note  when  the 
middle  point  of  each  is  made  a  node ;  in  which  case  the  note 
will  be  that  due  to  a  length  which  is  any  submultiple  of  \l. 

6.  If  a  body  hang  by  a  string,  and  through  any  point  of  the 
string  a  series  of  horizontal  lines  be  di-awn,  with  any  one  of 
which  the  body  may  be  rigidly  connected  and  perform  small 
oscillations  about  it,  the  time  of  oscillation  will  be  a  maximum 
about  a  line  at  right  angles  to  the  one  about  which  it  is  a 
minimum :  prove  this,  and  shew  how  to  find  the  position  of  these 
two  lines,  and  the  time  of  oscillation  about  any  other,  in  tenns 
of  the  times  about  these  two  and  the  angle  which  it  makes  with 
them. 

If  t  be  the  time  of  the  body's  oscillation  about  any  one  of  these 
lines  about  which  its  moment  of  inertia  is  Q^  we  have  the  relation 

« = ^^  (j4)'' 


318  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

where  M  is  the  body's  mass,  h  the  depth  of  its  centre  of  gravity 
below  the  horizontal  line  in  question.  Now  the  general  ex- 
pression for   Q  is 

Q  =  ^hn{x'  +  ^f  +  z'); 

therefore  if  we  make  the  horizontal  plane  through  the  lines  the 
plane  of  xy^  'S.hmz^  is  the  same  for  all  the  lines,  and  the  relative 
magnitudes  of  Q  for  the  different  lines  will  depend  upon 

But  this  expression  will  be  unaltered  if  we  project  every  particle 
of  the  body  upon  the  plane  of  xy.  Hence,  as  in  the  case  of  a 
plane  lamina,  we  shall  have  two  axes  in  the  plane  at  right  angles 
to  each  other,  for  one  of  Avlilch  Q  will  be  a  maximum  and  for 
the  other  a  minimum :  hence  the  first  part  of  the  proposition 
is  tnie. 

Also,  as  far  as  the  part  ^hn  (.c^  +  y"^)  of  Q  is  concerned,  we 
shall  have  the  usual  relation 

Q  =  Q,  co^'9  +  Q,^  sin'6', 

where  Q^^  Q^  are  the  maximum  and  minimum  moments,  and  6 
the  angle  which  the  axis  of  Q  makes  with  that  of  Q^ :  hence  the 
true  relation  between  Q^  Q^^  and  Q^  is 

Q  -  28m.z'  =  {Q^- ^Bmz^)  cos'0  +  {Q,-  ^Smz')  sm-*^, 

or   Q  =  Q^  cos'^^  +  Q^  sin^^,  as  before. 

Hence,  if  #,  t^^  t^^  be  the  times  about  the  lines  about  which 
Qi  Qit  Qii  ^1'®  the  moments  of  inertia, 

t  =  {t'^  cod'd  +  f.;  sln'^)i 

To  find  the  positions  of  the  lines  of  greatest  and  least 
moments  in  the  given  horizontal  plane. 

Let  the  direction-cosines  of  this  plane  referred  to  the  prin- 
cipal axes  thi'ough  the  point  where  the  string  pierces  the  plane 
be  ?,  m^  n]  and  let  Q  be  the  moment  about  the  line  whose 
direction-cosines  referred  to  the  same  axes  are  a,  /3,  7 ;  then  we 
are  to  have,  if  ^,  B^  C,  be  the  principal  moments, 

Q  =  Ad^  +  B^^  +  C<f  =  a  maximum  or  minimum 


1851.]  RIGID    DYNAMICS.  319 

subject  to  the  conditions 

a'  +  13'    +  7'^  =  1, 
and    /a  +  w/3  +  ny  =  0, 
which  last  is  the  equation  to  the  horizontal  plane. 

Differentiating  these  three  equations  with  respect  to  a,  /9, 
and  7,  we  find 

Aada  +  B^d^  +  C^^dy  =  0 (1), 

ada  +  ^dl3     +  rydy     =0 (2), 

and    Ida  +  md^    +  ndy     =0 (3). 

Using  the  arbitrary  multipliers  \  and  //.,  we  deduce  the  equations 

Aa  +  \a  =  fil (4), 

Bl3  -\-\^  =  fZ7n (5), 

Cy  +  \y  =  fjin (6). 

(4)  a  +  (5)  /9  +  (6)  7  gives 

Q  +  \  =  0', 

.-.    {A-Q)a  =fil  1 

{B-Q)^  =  f,m\ (A). 

{C-Q)y=jjLn} 

Hence  a^+  ^^+y^=l=  /."^  j^-^^  +  ^^.  +  (-^|-(7), 

and 

1  r  m^  w' 

_(?«  +  ,„^  +  ,,^)  =  0  =  ^—^  +  ^3^  +  ^^, 

which  gives  a  quadratic  for  the  detenmination  of  Q^  and  Q^. 
The  substitution  of  the  value  of  yu.  from  equation  (7)  in  equations 
(A)  will  give  us  the  values  of  a,  /3,  7,  and  so  dctcnnine  the 
positions  of  the  lines  of  maximum  and  minimvun  moments. 

1851. 

1.  The  locus  of  an  axis  passing  through  a  fixed  point  of  a 
solid  body,  and  such  that  the  moment  of  inertia  round  it  of  the 
body  is  constant,  is  a  cone  of  the  second  order,  and  the  cones 


320  SOLUTIONS  OF   RENATE-HOUSE   PROBLEMS.  [1851. 

coiTesponding  to  different  values  of  the  constant  moment  have 
the  same  directions  of  circular  sections. 

The  expression  for  Qj  the  moment  of  inertia  about  an  axis 
whose  direction-cosines  referred  to  the  pi*incipal  axes  are  a,  /3,  7, 
in  terms  of  the  principal  moments  A^  B,  C,  is 

or  if  the  axes  of  A,  B,  (7,  are  those  of  .r,  ?/,  and  2:, 

Q  [x'  +f  +  z')  =  Ax'  +  By""  +  Cz\ 

the  equation  to  a  cone  of  the  second  order. 

Let  Aj  B,  Qy  (7,  be  in  descending  order  of  magnitude,  the 
above  equation  may  be  put  in  the  form 

{A-Q)x'  +  {B-Q)f  -  {Q-C)z'  =  0; 

and  if  2  =  mx  +  7iy  +  c 

be  the  equation  to  any  plane  which  cuts  the  cone  in  a  cu'cle, 
we  have  {Oregory''s  Solid  Geometry^  Art.  124) 

n  =  0, 

_  ^  \A-Q-[B-Q)\^ 

'A-B\i 

=  +  ' 


.B-CJ' 

which  shews  the  direction  of  the  circular  sections  to  be  inde- 
pendent of  Q. 

2.  Determine  the  motion  of  a  heavy  solid  composed  of  two 
equal  right  cones  placed  together  base  to  base,  and  which  rolls 
without  sliding  upon  two  intersecting  lines  inclined  at  equal 
angles  to  the  vertical,  the  common  base  of  the  cones  moving 
in  the  plane  which  bisects  the  angles  between  the  vertical  planes 
through  the  lines. 

We  shall  apply  the  principle  of  vis  viva. 
Let  CA^  CB  (fig.  109)  be  the  two  lines ;  and  at  the  time  t 
let  them  touch  the  two  cones  in  P,  P' :  let  GM  the  height  of  G 


1851.]  RIGID    DYNAMICS.  321 

the  centre  of  gravity  of  the  soHd  above  tlie  horizontal  plane 
through  (7=2;,  CM  =  x.  Then  if  r  be  the  distance  of  P  from 
the  axis  of  the  cone,  and  S  the  inclination  of  the  plane  A  CB 
to  the  horizon,  FN  the  height  oi  P  =  z  —  r  cosS,  or  if  CP  =  p, 

psiny  =  z  —  rcosS; 

.■.  2;  =  p  sin7  +  rcosS (1). 

Also,  if  z  MCN  =  £, 

X  =  CiV coss  —  pcoay  coss (2). 

Now  as  the  cone  rolls  along,  the  locus  of  P  on  the  cone  will 
be  a  cui've  like  the  dotted  curve  in  the  figure :  let  Bs  be  an 
element  of  the  length  of  this  curve  answering  to  the  rotation 
of  the  solid  through  a  small  angle  89^  then 

Ss  —  S  (rcoseca)  cosecyS (3), 

if  /3  be  the  inclination  of  either  rod  to  the  common  base  of  the 
cones,  or  2/S  the  inclination  of  the  rods  to  each  other :  also 

8  (rcoseca)  =  8[r0)  tan/9 (4) ; 

.*.  8p  =  -  8s  =  —  Sr  coseca  cosec/S  by  (3) ; 

.•.  p  =  (r^  —  r)  coseca  cosec/S, 

if  r^  is  the  radius  of  the  common  base  of  the  cones :  also  by  (4), 

r  —  r^  =  rStan^  sin  a, 

if  0  =  0  when  the  solid  touches  both  the  lines  at  (7; 

.•.   0  =  ( 1 -]  cotS  coseco- 

V         rj 

Hence,  by  (1)  and  (2), 

z  =  (>;|  —  r)  coseca  cosec/S  sm 7  +  ?-cosS (5), 

X  =  (r^  —  r)  coseca  cosec/S  C0S7  coss. 


Now  the  equation  of  vis  viva  gives  us 


322  .SOLUTIONS   OF   SliNATE-IIOlSE    PKOBLEMS.  [1851. 

where  «„  is  the  height  of  p  where  tlic  solid  starts  from  rest ; 
.',  {cosec^rt  cosec'''/9  cos'^y  cos'^'s  +  (coseca  cosec/9  sin 7  —  cosS)"'' 

Pi-:'      ,._  ,  (dry 


l'<      v^o  1  (dry 

+  —^  cotp  coseca}  I  ^-1 


=  2/y  [z^^  —  7\^  coseca  cosec/3  sin7 

+  r  (coseca  cosec)8  sin7  —  cos8)}, 

an  equation  which,  when  integrated,  will  give  lis  r,  and  there- 
fore also  X  and  2,  at  any  time  t. 

From  equation  (5)  it  appears  that  as  z  will  necessarily  be 
diminished  by  the  force  of  gravity,  if  the  solid  starts  from  rest 
it  will  roll  toward  C  or  from  it,  according  as 

coseca  cosec/3  sin7  >  or  <  cos 8. 

3.  Shew  that  the  diiference  of  the  moments  of  inertia  of 
a  body  round  two  axes  in  a  given  plane  which  are  equally 
inclined  to  a  fixed  line  in  the  same  plane,  is  proportional  to  the 
sine  of  the  angle  between  those  axes. 

Let  ^j,  Q^  be  the  maximum  and  minimum  moments  about 
lines  in  that  plane,  Q^  Q  the  moments  about  any  two  lines  in 
the  plane  making  angle  0,  0'  with  the  axis  of  the  moment  Q^ ; 
then,  by  Problem  6,  1850, 

<2  =  ()^cos'0  + <?^sin'0, 

.-.    Q-  Q  =  Q^  {co&'e-QOs'B')  +  Q,^  (sin"-"0-sin'^0') 

=  Q^  sin  {B'-B)  sin  {B  +  B')  +  Q,^  sin  {B  -  B')  sin (0  +  B') 

=  {Q,-Q,)sm{B-\-B')  sin {B'-B) 

=  ((?,-^Jsin2asin(0'-0), 
where  a  is  the  angle  the  fixed  line  makes  with  the  axis  of  Q^ 

oc  sin(0'-0), 

Gc  sine  of  the  angle  between  the  axes  of  Q  and  Q'. 


{     323 


HYDROSTATICS. 

1849. 

1.  A  plane  body,  one  of  the  edge3  of  Avhich  is  a  straight 
line,  is  immersed  in  water  so  as  to  have  this  straight  line  coin- 
cident with  the  surface;  shew  how  the  depth  of  the  centre  of 
pressure  may  be  deduced  from  observation  of  the  time  of  a  small 
oscillation  in  vacuum  of  the  body  about  its  rectilinear  side. 

WTien  the  body  is  immersed  with  its  plane  vertical, 

let  z  be  the  depth  of  the  centre  of  pressure  below  the  surface, 

•••  z gravity  , 

...  z  any  point  of  the  body 

Jgpz'dz 


Then  z'  = 


Jf/pzdz 


Mz       z  ' 
(where  h'  is  the  radius  of  gyration  about  the  straight  edge) 


the  length  of  the  simple  pendulum  when  the  body  makes  small 
oscillations  about  the  rectilinear  side.  Hence,  if  T  be  the  time 
of  small  oscillations. 


r=27r 

.'.Z^l^C,—^, 

the  formula  for  the  detcnnination  of  z  from  T. 

2.  A  cylinder  the  radius  of  which  is  a,  having  its  axis 
vertical  and  containing  incompressible  fluid  (density  /a),  re- 
volves  about   its   axis  with   an    angular   velocity    o)  =  [»-r, 

Y2 


324  SOLUTIONS   OF   SENATE-HOUSE    PR(lBLEMS.  [1849. 

n  being  >  1  ;  a  sphere  (density  p'),  whose  radius  is  also  «,  on 
being  put  into  the  cylinder,  is  supported  in  such  a  position  that 
it  touches  the  free  surface  at  its  vertex :  shew  that 


p        \         n 
The  equation  of  equilibrium  is 

volume  of  fluid  displaced  =  —  .^Tra" (1). 

r 

Fig.  110  shews  the  position  of  the  sphere  in  the  fluid,  the 
dotted  line  representing  the  continuation  of  the  section  by  the 
plane  of  the  paper  of  the  free  surface. 

Let  F,  the  vertex  of  the  free  surface,  be  the  origin  of 
coordinates,  the  axis  of  the  cylinder  that  of  2,  and  r  the  distance 
of  any  point  from  it.  Then  the  equations  to  the  free  surface 
and  that  of  the  sphere  are 

2  2a 

CO    ^  n     ^ 

and  r'"'  =  2az  —  s'"* : 

therefore  if  z'  be  the  height  above  V  of  the  circle  of  intersection, 

or  z'  =  2(7  f  1 

Hence  the  volume  of  the  fluid  displaced 

=  irj^{2az-z')dz--.- 


TT  -I «  (  1   —  -]  z"^  —  !«'■ 


3' 


=  W(l-l){a(l-i)-Sa(.-l 


1849.] 

therefore  from  f  1 ' 


IIYDUOHTATrcS. 


^.|W  =  ..a^(l-^,, 


325 


or 


-  =  ('--)' 

>        V       nj 


3.  If  X,  F,  Z^  be  the  forces  acting  at  a  point  [xyz)  of  a  mass 
of  heterogeneous  fluid  in  equilibrium,  and  Xdx  +  Ydy  +  Zdz^ 
be  not  a  perfect  differential,  then  the  pressure  and  density  will 
be  constant  thi'oughout  the  curves  of  which  the  differential 
equations  are 

dx dy dz 

d]^_d^~  dZ _dX~  dX  _dY' 
dz       dy       dx       dz       dy       dx 

Let  p^  p  be  the  pressure  and  density  at  the  point  (xyz) ; 
])  +  dp  the  pressiu'e  at  the  point  [x  +  dx^  y  +  dy^  z  +  dz)^  then 

dp  =  p  {Xdx  ^-Ydy  +  Zdz) (1). 

In  order  that  this  equation  may  hold,  and  therefore  equi- 
librium be  possible,  we  must  have  the  right-hand  side  of  this 
equation  a  perfect  derivative  of  tkree  independent  variables, 
or  we  must  have 

dp  Y     dpZ 


dz 

dpZ 
dx 

dpX 
dy 


dpX 

dpY 
dx   ' 


or 


ldY_  _dZ\  ^  ^dp  _  ydp^ 
'^  \dz       dy)  dy  dz 

dZ      dX\  __   Y  ^P       ydp 
dx       dz  J       ^    dz  dx 


(A). 


dX 


(' 

\  dy       dx 


dY\  _  Y^  _  x^^ 


dx 


dy. 


326  SOLUTIONS   OF   SKNATE-HOUSE    PliOHLEMS.  [1849. 

^lultiplying  these  equations  in  order  by  A',  Y^  Z,  and  adding, 
we  get 

as  the  condition  wliicli  the  forces  X,  F,  Z^  must  satisfy  in  order 
that  they  may  be  able  to  produce  equilibrium. 

Now  let  dx^  dfy,  dz,  in  the  expression  (1)  for  f^>,  be  such  that 

dx fh/ dz  ,  . 

d_Y_dZ^(lz'_t(X~clX_dY ^'' 

dz       dy       dx       dz        dy        dx 

then  dp  will  be  the  variation  of  p  as  we  pass  from  one  point  to  the 
adjacent  point  of  any  of  the  curves  of  which  these  are  the  dif- 
ferential equations.     Now,  combinmg  (l)  and  (3),  we  get  by  (2) 

dp  =  0, 

wherefore  p  is  constant  along  the  curves  whose  differential 
equations  are  (3). 

Also  from  equations  (A)  we  may  put  equation  (3)  in  the  fonn 

dx  dy  dz 

dy  dz  dz  dx  dx  dy 

dy  =.  r(x'^  -  Zf 

\      dz  dx 

dz  =  r{Y^i-xf\: 
V      dx  dyj 

and  multiplying  these  equations  by  y- ,  y- ,  -^  ,  respectively, 
we  get  '         ^ 

dp  =  -fdx  +  -^  dy  +  ~dz  =  0: 
dx  dy  dz 

or  p  is  also  constant  along  these  curves. 


HYDROSTATICS,  327 

1850. 

1.  Three  equal  cylinders  arc  placed  in  contact  upon  a 
horizontal  plane,  sufficiently  rough  to  prevent  sliding:  find 
how  much  water  must  be  poured  into  the  space  between  the 
cylinders,  in  order  to  disturb  the  equilibrium. 

Let  h  be  the  depth  of  the  water  poured  in  when  each 
cylinder  is  on  the  pomt  of  turning  about  a  tangent  line  to  its 
base,  in  which  case  the  water  will  run  out  between  the  cylinders. 

Now  the  moment  of  the  fluid  pressures  upon  each  cylinder 
about  the  tangent  line  to  the  base  about  which  the  cylinder 
would  begin  to  turn,  is  the  same  as  the  moment  of  the  fluid 
pressures  on  a  vertical  rectangle  of  height  h  and  breadth  equal 
to  the  radius  [r]  of  each  cylinder  about  its  base* 


=  /'  I    gp  [h  —  z)  zdz 


Now  this  must  equal  the  moment  of  the  weight  [Mg]  of 
each  cylinder  about  the  same  line  or  Mgr^ 

is  the  required  height. 

2.  All  space  being  supposed  filled  with  an  elastic  fluid  whose 
volume  at  a  given  density  is  known,  the  particles  of  which  are 
attracted  to  a  given  point  by  a  force  varying  as  the  distance : 
find  the  pressure  on  a  circular  disc  placed  with  its  centre  at  the 
centre  of  force. 

Let  fi  =  absolute  force  of  attraction  at  distance  unity ;  the 
attractions  A',  Y,  Z^  parallel  to  the  axes  at  the  point  [xyz]  are 


•  For  this  is  the  natiire  of  the  sectiuii  of  each  cylinder  supposed  of  a 
height  /(,  made  by  a  plane  through  the  lines  of  its  contact  with  the  other 
cylinders. 


SOLUTIONS   OF   SENATE-HOUSE    PKOULEMS.  [1850. 

— /i.r,  — /i^,  —  t^^i  tlic  centre  of  force  being  origin:    hence 
dji  =  p  [Xdx  +  Ydy  +  Zdz) 

=  —  fip  [xdx  +  ydy  +  zdz)  =  —  yuprdr    if  r'-  =  x^-  +  ?/■'  +  s'' ; 
and  p=kp\ 

.'.  -^  —  —  ukrdr ; 


To  detennine  C,  wc  have 


p  =  kp  =  CkB-^'^""  • 

BM  =  mass  contained  between  two  con- 
secutive spheres  having  C  for  cen- 
tre, radii  r  and  r  +  Sr 

=  47rpr'S?-  =  A7rCk.rh'^'""-'Br', 

.  3/  =  whole  mass,  and  therefore  known, 


ATvCk 


Let  ^jfikr^  =  2, 


and  r'Wr  = z^dz  • 

.-.  M=4.7rCk-=—  /    sW^: 


and 


/%^s-v.  =  r(f)  =  ir(i) 


and  M  is  known ;  hence  G  is  also  known. 

Hence,  if  P  be  the  pressure  on  the  annulus  (radius  a)  we  have 
8P  =  27rrBr.j) 


1850.J  HYDROSTATICS.  329 

from  r  =  0)  /^'^ 


3.  A  hollow  cylinder  is  filled  with  inelastic  fluid  and  made 
to  revolve  about  a  vertical  axis  attached  to  the  centre  of  its 
upper  plane  face  with  a  velocity  sufficient  to  retain  it  at  the 
same  inclination  to  the  axis.  Find  at  what  point  of  the  face 
a  hole  might  be  bored  without  loss  of  any  fluid. 

Let  ft)  be  the  angular  velocity  of  rotation :  then,  if  the  fluid 
were  contained  in  an  open  vessel,  the  latus-rectum  (Z)  of  the 

generating  parabola  of  the  free  surface  would  be  -^.     Now 

since  it  is  supposed  by  the  question  that  there  is  a  point  in  the 
upper  plane  face  where  the  pressure  of  the  fluid  is  zero,  it  is 
manifest  that  the  face  touches  the  above  free  smiace  at  this 
point.  This  point  will  evidently  lie  in  the  diameter  of  the  face 
most  inclined  to  the  horizon,  at  a  distance  r  suppose  from  the 
centre  of  the  face.  Let  a  be  the  inclination  of  the  face  to  the 
vertical,  li  the  distance  of  the  vertex  of  the  supposed  free  smiiaee 
above  the  centre  of  the  face,  the  equation  to  the  free  surface  is 

and  for  a;,  y  we  may  write  r  cos  a,  r  sin  a, 

.-.  r^  sin'^a  =  Z  (r  cos  a  —  h) : 

the  roots  of  this  equation  are  equal, 

, ,  cosa 
.-.  r  =  \l  ^-^  . 
sm  a 

g     cos  a 


to     sm  a 


4.    A  mass  of  inelastic  fluid  is  contained  between  three  co- 
ordinate planes,  each  of  which  attracts  with  a  force  which  varies 


330  SOLUTIONS   OF   SKNATE-HOUSE    PROBLEMS.  [1850. 

as  tlic  distance,  and  the  absolute  forces  of  attraction  ytt,,  yu,.^,  fx^^ 
are  in  harmonic  progression.  Half  an  ellipsoid  is  fixed  with 
its  plane  surface  against  one  of  the  coordinate  planes,  and  its 
surface  touching  the  other  planes ;  its  axes  being  parallel  to  the 
coordinate  axes  and  proportional  to  //-,"*,  yu,./*,  fi'^.  If  there 
be  not  sufficient  fluid  quite  to  cover  the  ellipsoid,  the  uncovered 
part  will  be  bounded  by  a  semicircle. 

The  attractions  X,    F,  Z^  parallel  to  the  axes  arc  -  yu-jir, 

.'.  flp  =  Xdx  +  Y(hj  +  Zch     (if  p  =  unity) 
=  -  {H'.sccIx  +  fi_^ydy  +  fi^zdz) ; 
therefore  the  equation  to  the  free  surface  is 

fM^x^  +  fi^^y'^  +  fi^z^  =  a  constant  =  C  suppose (1). 

The  equation  to  the  ellipsoid,  if  it  be  bisected  by  xz^  is 

ti,[x-af  +  ^.y  +  fJi,{z-cy  =  C (2). 

(1)  —  (2)  gives  for  the  plane  of  intersection 

2fjb^ax  +  2/Zg02;  =  a  constant  =  2  ( C')^  A  suppose  ; 

•••  M'l^x  +  /*3*^  =  -4  (3), 


since   a  —  1       ,  ,      <-  —  i  — 

(3)  may  be  put  in  the  fonn 

fj,/'  =  A'  -  2Afi^^x  +  fi^x^ ; 
subtracting  this  equation  from  (1)  gives 

2/A^a?"  +  fJ^,f  =  2Afiix  -  A^  +  C (4), 

the  equation  to  the  projection  on  {xy)  of  the  curve  of  intersection. 
Let  <f)  equal  the  angle  at  which  (3)  is  inclined  to  xi/ ; 

and    cos"rf>  =  — — —  =  -^ 
since  /a,,  /w,^,  ^u,.^,  arc  in  hannonic  progression. 


tan0  =  I'-i 

V, 


1850.]  HYDUOSTATICS.  331 

This  equation,  taken  with  (4),  shews  that  tlie  axes  of  the 
projection  on  xij  of  the  curve  of  intersection  parallel  to  x  and  y 
respectively,  are  in  the  ratio  of  cos0  :  1;  hence  the  curve  of 
intersection  must  be  circular,  evidently  a  semicircle,  whose 
diameter  lies  in  xz^  and  its  plane  pei'peudicular  to  xz. 

5.  A  rectangular  vessel  is  filled  with  fluid  of  twice  its 
weight,  and  placed  with  its  open  end  downwards  upon  a  hori- 
zontal plane,  which  is  then  made  to  revolve  round  each  side 
of  the  base  successively,  one  of  these  sides  being  greater  and 
the  other  less  than  three  times  its  height:  find  when  the  fluid 
will  begin  to  escape  in  each  case,  supposing  the  centres  of 
gravity  of  the  vessel  and  the  fluid  to  coincide. 

If  the  vessel  had  a  base  instead  of  being  opened  at  the  lower 
end,  the  moment  of  the  fluid  pressure  on  its  inside  about  any 
side  of  the  base  would  be  the  same  as  that  of  its  weight  acting 
at  its  centre  of  gravity :  hence,  when  the  vessel  is  open  at  the 
lower  end,  the  moment  of  the  fluid  pressm-es  about  a  side  of  the 
base  will  be  that  of  the  weight  acting  at  its  centre  of  gravity, 
minus  the  moment  of  the  fluid  pressures  on  the  plane  on  which 
the  vessel  rests. 

To  find  this  moment,  M  suppose.  Let  the  horizontal  plane 
be  supposed  to  have  been  turned  through  an  angle  a,  and  let  r 
be  the  distance  of  any  point  in  it  from  a  horizontal  line  in  it, 
at  the  same  height  as  the  highest  edge  of  the  vessel :  the  dis- 
tance of  the  edge  about  which  the  vessel  is  being  turned  will  be, 
if  a  be  this  edge,  h  the  other  edge,  and  h  the  height  of  the 
vessel,  h  -f  h  cot  a.     Hence 

M  =  i  f/pr  sin  a .  adr  [h  +  h  cot  a  —  r) 

J  Acota 

pb*hcoti 

=  gpa  sina  I  [{b  +  h  cota)  r  —  /•"}  dr 

J  A  cota 

=  gpabsma{^{b+hcota)  {b  +  2hcota)-^[b''+3bhcota+3h''Qot''a)] 
=  gpab  sina  {\l>^  +  \1d}  cota). 

Let  Whc  the  weight  of  the  vessel,  and  therefore  2  W  that  of 
the  fluid:  then,  when  the  water  begins  to  flow  out,  the  ni(»ment 


332  SOLUTIONS   OF   SENATE-HOUSE    TKOBLEMS.  [1850. 

about  tlic  edge  ct  of  all  the  forees  on  this  vessel,  including  its 
weight,  is  zero,  or 

3  W{^b  cos  a  —  ^h  sin  a)  —  M  =  0, 
or,  since  W  =  gpabh,     and  b  =  Snh  suppose, 

—  [Stik  cosa  —  h  sina)  —  S7ih  sina  {^.3nh  +  ^h  cota)  =  0, 
or    3«  cosa  —  sina  —  n  {n  sina  +  cosa)  =  0  ; 

2n 
.'.  tana  =  -2 r; 

which  gives  the  value  of  a  when  the  two  values  of  n  are 
substituted,  one  >,  the  other  <  1.  In  both  cases,  however, 
a  is  <  45°. 


(     333     ) 


HYDRODYNAMICS. 

1848. 

1.  A  cylindrical  vessel,  with  its  axis  vertical,  is  filled  with 
fluid,  which  issues  from  a  great  number  of  small  orifices  pierced 
in  the  side :  find  the  surface  which  touches  all  the  streams  of 
spouting  fluid. 

This  surface  is  evidently  a  surface  of  revolution,  having  the 
axis  of  the  cylinder  for  axis.  Its  generating  curve  is  the  line 
which  touches  all  the  parabolic  jets  of  water  from  the  different 
orifices  in  the  same  generating  line  of  the  cylinder.  These  jets 
have  all  this  generating  line  for  axis,  and  a  common  directrix 
in  the  plane  in  which  they  lie,  viz.  the  horizontal  line  at  the 
level  of  the  surface :  for  the  velocity  of  efflux  is  that  due  to  the 
distance  from  this  line. 

Hence,  making  the  common  axis  and  directrix  axes  of  x 
and  y  respectively,  the  equation  to  the  jet  whose  point  of  efliux 
is  at  a  depth  h  is 

f  =  Ah{x-h). 

To  find  the  line  which  this  cuiwc  always  touches,  differentiate 
with  respect  to  /*,  considering  x^  y  constant ; 

.-.  0  =  a;  -  27*, 

and  eliminating  ^, 

f  =  'lx.\x, 

or     y  =  x\ 

the  equation  a  straight  line  through  the  origin,  inclined  to  the 
vertical  at  an  angle  of  45°.  Hence  the  surface  required  is  a 
right-angled  cone  placed  on  the  cylinder  in  an  inverted  position. 

2.  A  closed  vessel  is  filled  with  water,  containing  in  it  a 
piece  of  cork  which  is  free  to  move ;  if  the  vessel  be  suddenly 


334  soi.rTioNs  uF  senate-house  prohlems.         [1849. 

moved  forwards  by  a  blow,  shew  that  the  eork  will  shoot  for- 
wards relatively  to  the  water. 

Suppose,  for  an  instant,  the  eork  removed,  and  Its  plaee 
occupied  by  solidified  water ;  when  the  blow  is  stnick  this  mass 
of  solidified  water  will  instantaneously  receive  a  velocity  V  equal 
to  that  of  the  surrounding  water,  and  the  impulse  on  it  will  be 
MV^  if  M  be  its  mass.  But  when  the  cork  is  in  the  place  of 
this  solidified  water,  the  impulsive  actions  on  it  of  the  suiTOund- 
ing'  fluid  will  be  the  same  as  they  were  on  the  solidified  water, 
and  therefore  the  impulse  on  it  will  be  the  same.  But  the  cork 
is  lighter  than  the  same  volume  of  solidified  water,  and  therefore 
the  same  impulse  will  impart  a  greater  velocity  to  it,  or  the  cork 
will  move  forward  relatively  to  the  water. 

3.  A  closed  vessel  is  filled  with  water  which  is  at  rest,  and 
the  vessel  is  then  moved  in  any  mamier:  apply  the  principle 
of  the  consei'vation  of  areas  to  prove  that,  if  the  vessel  have  any 
motion  of  rotation,  no  finite  portion  of  the  w^ater  can  remain 
at  rest  relatively  to  the  vessel. 

The  principle  of  consei'vation  of  areas  about  any  axis  must 
apply  to  the  whole  mass  of  water.  But  if  any  portion  of  the 
water  remain  at  rest  relatively  to  the  vessel,  we  may  suppose 
it  to  become  solidified  and  rigidly  attached  to  the  vessel  without 
altering  the  motion  of  any  particle  of  the  water :  but  in  this 
case  it  is  evident  that  the  principle  of  the  couseixation  of  areas 
about  any  axis  must  also  apply  to  the  part  of  the  -water  not 
solidified ;  consequently  it  must  also  apply  to  the  solidified  poi- 
tion  of  the  water  which,  since  the  water  is  originally  at  rest, 
can  therefore  have  no  motion  of  rotation,  which  is  absurd  if. 
the  vessel  have  any  motion  of  rotation.  Therefore,  if  the  vessel 
have  any  motion  of  rotation  there  cannot  be  any  finite  portion 
of  the  water  which  remains  at  rest  relatively  to  it. 

1849. 

1.  Supposing  the  effect  of  friction  in  the  case  of  aerial 
vibrations  in  a  tube  of  uniform  bore  to  be  the  production  of 


1849.]  HYDRODYNAMICS.  336 

a   retarding  force    on    each    particle    equal    to   /  x  velocity, 
prove  that  the  equation  of  motion  will  be  satisfied   by  taking 

cs'^^'  sm  —  ■{  a 


f  1  —  Y— ^  j  t  —  xl  as  the  type  of  the  vibrations. 


Let  X  be  the  coordinate  of  any  particle  at  rest,  oc  +  ^  its 
coordinate  when  displaced  at  time  t ;  then  the  equation  of  motion 
will  be 

df  ~      clx'      J  dt  ' 
Now,  if  we  assume 

P  =  cr^^'  sin  —-  (nat  —  x),    where  7?'^  =  1  —    '.  .,   ., , 

^  \     ^  '^'  16(1-77    ' 

we  have 

-~  +/f  =  cz'^-^'  ■!  — T—  cos  —  {7iat - ^)  +  ^  sin  —  {nat  —  x)[', 

,„  47r''^fl^    .    27r  .  , 

=  —  c£ -•"       ■;     sm  —  (wa^  —  ic), 

by  substitution  of  the  value  of  n. 

Ai  '2  ^"1  47r'''(t''^       ,,,    .     27r  ,  , 

Also  a  -T-T,  = ^-5-  «?£  *-^^  sm  ^{nat  —  x)\ 

dx-  X  X   ^  ' 

' '   df  '^•^  dt  dx' 

^^      d''^  d"^         d^ 

and  the  equation  of  motion  is  satisfied. 

2.  Steam  is  nishlng  from  a  boiler  through  a  conical  pipe, 
the  diameters  of  the  extremities  of  which  are  D  and  d  respec- 
tively :  prove  that  if  V  and  v  be  the  coiTesponding  velocities 
of  the  steam, 

r  =  T   —  £  =* 


336  SOLUTIONS   OF   SEXATE-IIOUSE    PROBLEMS.  [1851. 

where  k  is  the  pressure  divided  by  the  density,  and  supposed 
constant.  The  motion  may  be  supposed  to  be  that  of  a  fluid 
diverging  from  a  centre,  the  centre  being  the  vertex  of  the  cone, 
of  whicli  tlic  pipe  fonns  a  portion. 

Let  J)  and  p  be  the  pressure  and  density  at  the  distance  ?• 
from  the  centre  of  motion  at  the  time  t,  when  the  velocity  at 
that  point  is  u ;  then,  since  the  motion  is  wholly  radial,  its 
equation  is 

1  dp  du  du       .         .  .       .  -  ,  . 

~  'j~  —  ~  ~ji  ~  ~  '^zr  y  smce  the  motion  is  steady  ...  (1). 

Also    2)  =  hp (2). 

The  consideration  of  continuity  gives  the  equation 

?//)r^  =  constant, 

or     ^ipr^  =  constant (3). 

From  (1)  and  (2), 

k  dp  du 

p  dr  dr  ^ 

or     k  \o^p  =  C  —  ^ii\ 
Let  P,  p  be  the  pressures  at  the  two  extremities  of  the  pipe, 


But  from  (3), 


^•log 

P 

w- 

or 

P  _ 

P  ~ 

£    2*      . 

P 

vd' 

P~ 

VB'' 

.    V  = 

'    d' 

185L 

L  If  a  regular  homogeneous  tetrahedron  be  placed  in  any 
position  whatever  in  a  fluid  whose  density  varies  as  the  depth, 
shew  that  when  the  resistance  of  the  fluid  is   neglected,  the 


1851.]  HYDRODYNAMICS.  337 

7A4 


tetrahedron  will  make  vertical  oscillations  in  the  time  2-77 

h  being  the  depth  of  the  centre  of  gravity  of  the  tetrahedron  in 
the  position  of  equilibrimn. 

We  shall  first  shew  that  the  centres  of  gravity  of  the  tetra- 
hedron and  of  the  displaced  fluid  are  in  the  same  vertical,  what- 
ever position  the  tetrahedron  occupies  in  the  fluid. 

Let  the  centre  of  gravity  of  the  tetrahedron  be  at  a  depth  z 
below  the  sui*face,  and  be  taken  as  origin  of  rectangular  co- 
ordinates [xyz)^  the  latter  vertically  downwards.  Let  the  den- 
sity at  a  depth  z  below  the  surface  be  fiz :  the  density  at  the 
point  xyz  will  be  ix[z'  +  z)  =  c  +  fiz  suppose. 

Let  r,  y,  i,  be  the  coordinates  of  the  centre  of  gravity  of  the 
displaced  fluid ; 

.-.  (mass)  X  =  JfJ{c  +  fiz)  xdxdydz : 
but 

JJJxdxdydz  =  0, 

because  the  centre  of  gravity  of  the  solid  is  origin,  and 
JJfxzdxdydz  =  0, 

because  every  system  of  rectangular  axes  through  the  centre  of 
gravity  of  a  regular  solid  is  2i  2)'>'incipal  system  ; 

.'.  ^  =  0,     and  similarly  y  =  0 : 

hence  the  centre  of  gravity  of  the  fluid  displaced  lies  in  the 
vertical  line  through  that  of  the  tetrahedron. 

Thus,  in  whatever  position  the  tetrahedi'on  be  originally 
placed,  its  centre  of  gravity  will  move  m  a  vertical  line,  and 
make  finite  oscillations  in  that  line. 

The  force  acting  downwards  on  the  solid  at  any  time 

=  the  weight  of  the  solid  —  weight  of  fluid  displaced 

=  g(TV  -g  fjj{c  +  fiz)  dx  dy  dz 

(if  V  be  the  volume  of  the  tetrahedi'on,  o-  its  density) 

=  go-  V  —  gc  F, 

J5. 


""'  -df+''^'=^^ 


338  SOLUTIONS  OF   SENATE-HOUSE   rilOBLEMS.  [1851. 

since  the  centre  of  gravity  is  the   origin  of  coordinates,  and 

therefore 

fjjz  dx  dy  dz  =  0. 

Hence  the  equation  of  oscillating  motion  is 

since  c  =  /xz'j 

and  the  time  of  an  oscillation 

Now  A,  the  depth  of  the  centre  of  gravity  in  the  position,  is 

d'^z 
the  value  of  z  in  the  above  equation,  when  —p^  =  0 ; 

or     A  =  —  , 
and  the  time  of  an  oscillation 

This  proposition  is  equally  true  of  any  homogeneous  regular 


solid. 


(    339 


GEOMETRICAI.  OPTICS. 

1848. 

1.  Compare  the  brightuess  of  the  Earth  as  seen  from  Venus 
with  the  brightness  of  Venus  as  seen  from  the  Earth,  supposing 
the  sizes  and  reflecting  powers  of  the  two  bodies  equal. 

Let  Sy  E,  Vy  (fig.  Ill)  be  the  respective  positions  of  the  Sun, 
and  the  centres  of  the  Earth  and  Venus,  at  the  time  when  their 
brightness  is  to  be  compared. 

From  E  di-aw  the  straight  lines  JSa,  Eb  perpendiciilar  to  ES 
and  EV  m  the  plane  of  the  ecliptic,  and  similarly  T  c,  Vd  per- 
pendicular to  VS  and  VE:  the  part  of  the  Earth  seen  from 
Venus  will  be  contained  between  planes  pei'pendicular  to  the 
ecliptic  through  Ea^  Eb ;  and  the  part  of  Venus  seen  from  the 
Earth  between  the  planes  Vc^  Vd. 

Let  Q  be  the  quantity  of  light  that  falls  upon  a  imit  of 
the  sm*face  of  Venus  which  has  the  Sun  in  its  zenith.  To  find 
the  quantity  of  light  reflected  to  the  Earth  from  ajiy  element 
8S  of  the  sm-face  of  Venus. 

Let  the  latitude  and  longitude  of  the  element  B8,  referred 
to  the  Sun  as  origin,  and  plane  of  the  ecliptic  as  plane  of 
longitude,  be  6  and  (f).  The  quantity  of  light  reflected  to  the 
Earth  from  SS  will 

=  QBS  X  cosine  z.  d.  of  Sun  x  cosine  z.  D.  of  Earth 
=  QBS.cos9cos(li.cosecos{V-(f>),  F=  l  SVE, 

=  Q .  r'^  Bd  cos^  B(f) .  cos^O  cos<f)  cos(  V—  ^), 
if  r  =  radius  of  the  Earth  or  Venus. 

Hence  the  whole  light  reflected  to  the  Earth 
=  Qr'JJcoB'd  cos</)  cos(r-<^)  dd  d<f) 
=  i<3r7/cos''^(co3F+cos(F-2<^)}  dd  d(f> 
=  ^Qr'Jcofi'e{cosV.<f>-^sm{V-  2<^)  +  Oj  <7^: 

z2 


340  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

from  <f>  =  {V—  ^7r]] 
to       </)  =  Itt 

=  ^  Qr""  Jcos'e  [cos  r  (tt  -  F)  +  ^  sin  (tt  -  T^) 

=  ^  Qr-'  {(tt  -  F)  cos  V+  sin  F}  /cos'(9  tZ^ 
=  ^Qr^  {(tt-  F)  cosF+  BinF}/(cos3^  +  3  cos^)  d0: 
from  ^  =  —  ^tt"! 

to  0  =  +  i-TTJ 

=  |(2^.2((7r-F)cosF+sinP"|. 
Similarly,  the  whole  light  reflected  from  the  Earth  to  Venus 
=  f  Q'r'  {(tt  -  E)  cosE+  sin^}, 

when  Q'  is  the  quantity  of  light  that  falls  upon  a  unit  of  the 
Earth's  sui'face  which  has  the  Sun  in  its  zenith ; 

and  the  required  ratio 

_8V^   (tt-E)  cosE+s'mE 
~  SE' '  (tt  -  F)  cos  F+  sm  F  • 

2.  Find  the  geometrical  focus  of  a  pencil  of  rays  refracted 
through  a  hollow  glass  sphere,  whose  external  and  internal  radii 
are  ?•,  r'  re^ectively. 

Let  ti,  Vj,  ^25  ^3)  ^^^  ^3  be  the  distances  from  the  centre  of 
the  sphere  of  the  foci  before  the  1^'  and  after  the  P',  2°^*,  3'^'^,  and 
4"^''  refractions  respectively. 

Then,  by  the  common  fonnula, 

i  =  _^i:zi  +  ^ (,), 

1-  -'LLI.      1    i 


1848.1  GEOMETRICAL  OPTICS.  341 


or  t^tzl  +  1. (5), 

V,  r  v^ 

i=^+^ (3), 

d     ^       At"  ^       1    1 
or   :^  =  -  C +  _ (4  . 


Adding  (1),  (2),  (3),  and  (4), 

1       1\    .   At 


and  for  glass,  /i  =  f  ; 


1    Q  n    i\    1 

which  determines  v,. 


V,      3  U'       rj  "^  w ' 


3.  Light,  proceeding  from  a  given  point  P,  suffers  any  num- 
ber of  reflections  and  refractions :  if  consecutive  rays  of  a  given 
colour  come  out  parallel,  in  the  direction  determined  by  angular 

coordinates  ^,  (f),  shew  that  -y-  ,  -^  may  be  obtained  by  dif- 
ferentiating as  if  the  differently  coloured  rays  which  severally 
come  out  parallel  to  their  consecutives  started  from  P  in  the 
same  direction. 

Application.  In  the  case  of  the  rainbow  of  the  2^^^  order, 
given 

D  =  pjT  +  2(f)  —  2{2)-\-  1)  (^',     sin^  =  /i  sin0', 

find  the  order  of  the  colours. 

In  general,  if  ^,  ^  be  the  coordinates  upon  cniorgencc  of  the 
ray  whose  coordinates  as  it  proceeded  from  F  were  0\  <f>\  and 
if  /A  be  the  refractive  index  of  the  ray, 


342  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1848. 

Now,  If  0  and  (f)  be  the  coordinates  of  the  rays  of  refractive 
Index  fji  whicli  come  out  parallel  to  their  consecutives,  0',  <f>' 
may  be  foiuid  in  terms  of  /x.  from  the  equations 

d6' '^  d<f>"  dd'  ~    ' 

d^      d^    ^'  _ 
dS'      dii>' '  dB'        ' 

those  consecutive  rays  being   supposed  to  come   out  parallel 
whicli  before  Incidence  lie  In  a  plane  defined  by  the  particular 

value  of  -^  . 
du 

Now,  supposing  0  and  </>  to  retain  the  particular  meaning 

assigned  to  them  above,  since  &  and  <^'  are  now  fimctlons  of  /t, 

dB      (dd       dO    d<f>'\  dO'      dO 


dfi      [dO'"^  dcf>"de')  dfi'^  dfi' 

d(f>  _  /d(f)       d(f)    d<}>'\  dO'      # 
dfi       \dd'       d(f)'  '  dd'  J  dfi       dfx  ' 

The  first  tenns  In  these  expressions  for  -7-  ,  -^  correspond 

to  the  variation  of  the  direction  of  emergence  due  to  the  va- 
riation of  the  direction  of  Incidence  ;  the  second  tenns  coiTcspond 
to  the  variation  of  the  direction  of  emergence  due  to  the  va- 
riation of  /A  In  the  differently  coloured  rays.  But  we  have  seen 
that  the  first  tenns  are  each  equal  to  zero ;  hence  the  whole 
variation  of  6  and  <^  is  due  to  the  vaiiatlon  of  fj,  In  the  dif- 
ferently coloured  rays,  or  we  may  obtain  -7- ,  ■—  by  differen- 
tiating as  if  the  differently  coloured  rays  which  severally  come 
out  parallel  to  their  consecutives  started  from  P  in  the  same 
direction. 

In  the  application  to  the  case  of  the  rainbow  all  the  incident 

rays  are  parallel ;  we  may,  however,  differentiate  for  -^  ,  -^ 

as  if  all  the  angles  of  incidence  of  rays  which  come  out  parallel 
to  their  consecutives  were  equal. 


1848.]  GEOMETRICAL  OPTICS.  343 

Here  each  ray  which  comes  to  the  eye  moves  in  the  same 
plane;  and  since  the  rays  incident  upon  the  raindrop  are  all 
parallel,    the  angular  coordinate   after  emergence    will  be  i), 

the  deviation.     Hence,  to  find  -y-  we  differentiate  i),  consider- 


dD  „,        ^,  d<i>' 


ing  0  constant; 

and     s,in<f)  =  fi  sin0' ; 

...  ,,  dd) 

.'.  0  =  8m9  +  /J,  C0S9  -J-  ) 

and   -J—  =  —^ tan0   is  positive : 

hence  the  red  rays  which  come  out  parallel  to  their  consecutives 
will  be  more  deviated  than  the  violet  rays  which  come  out 
parallel  to  their  consecutives. 

It  only  remains  to  find  in  which  direction  the  rays  which 
form  the  rainbow  have  been  deviated.  To  ascertain  this,  we 
must  differentiate  D  with  respect  to  </>,  considering  fi  constant, 
and  put 

dD      ^ 

d^  =  '- 

This  equation  will  give  us  a  value  of  ^,  which  substituted  in  D 
will  determine  the  amount  of  deviation  of  the  rays  of  the  re- 
fractive index  /t,  by  which  the  corresponding  part  of  the  rainbow 
is  seen  :  let  this  value  be 

D  =  2mir  -f  i/r : 

then,  if  a^  be  <  tt,  the  deviation  at  the  first  refraction  will  be 
towards  the  eye,  and  the  red  rays  will  appear  on  the  inside  of 
the  arch :  if  i/r  be  >  tt,  the  deviation  at  the  first  refraction  will 
be  from  the  eye,  and  the  red  rays  will  appear  on  the  outside  of 
the  arch. 

4.  Every  diameter  {d)  of  the  extreme  boundary  of  a  sphe- 
rical reflector  subtends  a  right  angle  at  C  the  centre  of  the 
sphere:  supposing  parallel  rays  (inclined  at  an  angle  a  to  the 


^44  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1848. 

axis  of  the  reflector)  to  be  incident  upon  every  point  of  the 
extreme  boundary,  shew  that  the  section  of  the  reflected  pencil 
made  by  a  plane  passing  through  C  and  peqiendicular  to  the 
axis,  will  be  an  hyperbola,  whose  axes  are  d  and  J  cot  a. 

Let  CO  (fig.  112)  be  the  axis  of  the  mirror  AOB-^  let  Or, 
parallel  to  the  direction  of  incident  rays,  be  taken  for  axis  of  x ; 
Cy,  pei-pendicular  to  it  in  the  plane  OCx^  for  axis  of  y,  and  Cz 
pei-j^endicular  to  Cb,  Cy  for  axis  of  z. 

Let  Cx  pierce  the  miiTor  at  the  point  o:  Join  P,  any  point 
in  the  bomidaiy  of  the  miiTor  by  arcs  of  great  circles,  with 
0,  o :  call  Po,  PoA,  0  and  </> :    OF  will  be  45°. 

The  ray  reflected  from  P  will  pass  through  the  point  D 
of  Co,  such  that  the  perpendicular  from  D  upon  CF  bisects  CP, 
t.  e.  through  the  point  {^a  sec  6^  0,  o) :  also  the  coordinates  of  P 
are  a  cos^,  a  smd  cos^,  a  ski 6  s'm(f>.  Hence  the  equation  of 
the  reflected  ray  is 

X  —  ia  sec6  ii  z  ,  . 

a 1 a  = ^a         ±  =  — •     a    • — T  =  ^  SUppOSe...(l). 

aeost/  — ^a  secc^      asmc^cosip      asmc^sm9  ^^  ^ 

Also,  from  the  triangle  OPo, 

cos45°  =  cos^  cosa  —  sin^  sina  cos(^ (2). 

Our  object  will  be  to  eliminate  ^,  <^  from  these  equations, 
and  find  the  relation  between  r\  and  z  when  we  have  written. 

a;  =  ?;  sin  a, 
y  =  7]  cosar 

the  equation  so  fonned  will  evidently  be  the  equation  of  the 
curve  in  which  the  plane  through  C  cuts  the  surface  formed  by 
the  refracted  rays. 

From  (1), 

X  cos  0  —  la  =  a\  cos^d  —  \a\ ; 

.-.  aV  cos'a  -  a\x  cosB  +  ^x^  =  ^a^X{\  -  1)  +  Ix' ; 

.-.  a\  cos e  =  ^x+  {^d'\{\  -  1)  +  lx'}i (3)- 

Hence  equation  (2)  becomes,  multiplying  by  aX, 

-^a\  =  [^x  +  {^a^X{X-  1)  +  ^x^]  cosa  -  y  sina, 


1849.]  GEOMETRICAL  OPTICS.  345 

or  {|rt'''\(X—  1)  +  ^x"]*  cosa  =  -^  a\  +  y  sina  —  ^x  cosa, 

it" 

or   {^a''X(\—  1)  +  \7f  sin* a]*  cosa  =  ^  «^  +  ^  sina  cosa  ...(4) : 
hence,  squaring, 

\ci\  (A.  —  1)  cos'^a  =  \d'''}^  +  -^  ciXt]  sina  cosa ; 

.'.  aX  sin'''a  +  a  cos'^'a  =  —  %^i)  sina  cosa, 

and    aX  =  —  2-7;  cot  a  —  a  cot'^a. 

Again,  from  equations  (1), 

d'X'  Qm^e  =  f  +  z' : 

adding  this  equation  to  (3)'^, 

«V  =  ^x'  +  f  +  z'  +  K^(X-  1)  +  {ia'X(X-  1)  +  \x']^-x^ 

or  by  (4), 

ia'^X  (X  + 1 )  =  ^77^ + i77^cos'"'a + z^  ^ (  tu:  «^ + \'r]  sina  cosa  )  «  sina ; 

cosa  \2*  y  ' 

or  retaining  only  the  second  powers  of  t]  and  z^ 

Iff  cot'a  =  \'i]\\  +  cos'a)  +  z"*  -  tf  +  ^t^''  sin'a  +  &c., 

or   T\'  cot'^a  —  z''  =  &c.; 

shewing  that  the  required  locus  is  a  hyperbola,  the  ratio  of 
whose  axes  is  cot  a. 

Now  the  axis  in  the  plane  OCx  is  evidently  d:  hence  the 
axes  are  d  and  d  cot  a. 

1849. 

1.    A  ray  of  light  is  incident  upon  one  of  two  reflectors 

inclined  to  each  other  at  an  angle  -  ,  in  a  direction  parallel 

to  a  line  which  is  at  right  angles  to  their  intersection,  and 
bisects  the  angle  between  them :  supposing  the  intensity  of  a 
ray  reflected  at  an  angle  (j>  to  be  to  that  of  the  incident  ray 
as  e  cos(f>  to  1,  shew  that  the  intensity  of  the  ray  after  it  has 
suffered  n  reflexions  will  be  to  that  of  the  incident  ray  as  e" 
to  2"-'. 


346  SOLUTIONS  OF   SENATE-HOUSE   PROBLEAIS.  [1849. 

From  the  problem  on  p.  60  it  appears  that,  if  n  be  even,  the 
successive  angles  of  reflexion  are  complcmentaiy  of 

Itt      S    tt        7}  —  1    it      n  —  1    tt        Stt      Itt 
2'w'    2'n  '"       2      'w'    ~~2      'n"'2'n^    2n' 

Ilcncc  the  intensity  of  the  ray  after  n  reflexions  :  that  of  the 
incident  ray 


-"( 


.      1    TT        .      3    TT  .71—1      TtV      , 

sm  -  -  .  sm  -  -  ...  sm  — - —  .  -     :  1. 
2  n  2  71  2        nj 


Similarly,  when  n  is  odd,  this  ratio  is 

„  /  .    1  TT     .    3  TT         .    n    ttV    . 
e     sm  -  —  .  sm  -  -  . . .  sm  -  .  -     :  1. 
V       2  w  2  n  2     nJ 

Both  these  ratios  may  be  expressed  by  the  general  formula 

„.l7r      .37r  .2?i  —  Itt, 

e  siij  -  -  .  sm  -  -  . . .  sin  — :  1 : 

2  71  2  n  2       w        ' 

and  in  Hymers'  Tlieory  of  Equations^  Art.  22,  Ex.  20,  it  appears, 
by  making  ^  =  0,  that 

.      1    TT      .     3    TT  .      2«  -  1    TT  1 

sm  -  -   sm  -  -  . . .  sm  — - —  -  =  -^r=-,  , 
2  n         2  71  2        w      2    " 

whether  n  be  odd  or  even :  hence  each  of  the  above  ratios 

=  e"  :  2"-\ 

2.  A  transparent  medimn  is  bounded  by  two  parallel  planes ; 
the  refractive  index  is  constant  thi^oughout  any  plane  parallel 
to  the  bounding  planes,  but  varies  continuously  in  the  direction 
of  the  nonual  to  those  planes:  shew  how  to  find  the  path  of 
a  ray  of  light  tlu'ough  such  a  medium,  and  prove  that  in  passing 
through  a  section  for  which  the  refractive  index  is  a  maximmn 
or  a  mmimmn,  the  path  will  in  general  have  a  point  of  contraiy 
flexure. 

It  is  evident  that  the  path  of  any  ray  will  lie  in  one  plane : 
in  this  plane  take  two  lines,  one  perpendicular  to  the  bounding 
planes,  the  other  parallel  to  them  as  axes  of  x  and  y  re- 
spectively. 


1850.]  GEOMETRICAL  OPTICS.  347 

At  the  point  [xy]  let  the  inchnation  of  the  path  to  the  axis 
of  X  be  0,  and  let  the  refractive  index  be  yit:  at  an  adjacent 
point  [x  +  Sa-,  ?/  +  Sy),  let  these  be  ^  +  8^,  fi-\-  Bfi:  then,  by  the 
law  of  refraction, 

Bin<f>  =  — sm{(f>  +  S(f)) 


+  —  1  (sin^  +  COS080), 


or  0  =  cos  ^B(f)  ■+  —  sin</)S/i; 


therefore,  proceeding  to  the  limit, 


-J-  +  -  tan<^  =  0, 

,  ,  d6       1   da 
or  cot<i>  ~  +  -  -f-  =  0'. 
ax       fi  ax 

.'.  logging  +  log/i  =  constant  =  logO, 
.*.  sin©  =  —  , 


01*    1  +  I  ^-  I    —  I  Vy  /  ) 


dxV  _  ffM 
dy)   "  \G 
dy       f//.^^      ^-* 


which,  since  /i  is  a  known  function  of  a*,  is  a  diflferential  equation 
for  the  determination  of  the  path. 

7 

When  /x.  is  a  maximmn  or  minimum,  we  have  -y-  =  0, 

and  as  the  ray  passes  through  such  a  section,  its  path  usually 
suffers  inflexion. 

1850. 

1.   If  a  string  be  wrapped  round  a  glass  prism,  whose  section 
is  an  equilateral  triangle,  so  as  to  be  always  inclined  at  the  same 


348  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1850. 

angle  to  the  axis  of  the  prism,  the  portion  of  the  string  seen  by 
internal  reflexion  will  appear  to  be  parallel  to  the  portions  seen 
directly. 

Let  AB  (fig.  113)  be  a  portion  of  the  string  seen  directly, 
BC  a  portion  seen  by  internal  reflexion  at  the  surface  AC  oi  the 
prism :  and  first,  let  the  eye  be  In  such  a  position  that  the  small 
pencil  of  rays  by  which  any  point  of  BG  very  near  B  is  seen, 
shall  pass  through  some  point  of  the  surface  ABB'  very  near 
to  B\  a  point  in  the  edge  AB'.  Then,  if  the  eye  be  at  a  con- 
siderable distance  from  the  prism  so  that  the  axes  of  small  visual 
pencils  may  be  considered  parallel  to  each  other,  any  point  of 
BC  very  near  C  will  be  seen  by  a  small  pencil  which  passes 
through  the  surface  ABB'  at  some  point  very  near  C  a  point 
in  the  edge  BC.  Let  B'U^  G'E  be  the  directions  of  these  small 
pencils  upon  emergence.  Now  the  plane  CC'E  is  parallel  to 
the  plane  BB'E]  and  if  we  draw  a  plane  through  A  parallel  to 
these,  the  plane  BB'E  will  be  equidistant  from  the  other  two, 
since  the  prism  is  equilateral,  and  AB^  BC  equally  inclined  to 
its  axis;  hence  BC  =  AB'^  and  B'C  is  parallel  to  AB: 
hence  BC  appears  parallel  to  AB. 

Now  let  the  eye  be  moved  in  any  manner  without  approaching 
too  near  the  prism ;  it  may  easily  be  seen  that  the  locus  of  the 
points  where  the  rays  from  BC  to  the  eye  cross  the  plane  ABB' 
is  parallel  to  B'  C",  and  therefore  to  AB.  Hence  BC  will  always 
appear  parallel  to  AB;  and  the  same  may  be  shewn  of  any 
other  portion  of  the  string  seen  by  internal  reflexion. 

2.  A  rectangular  box,  at  the  bottom  of  which  is  a  plane 
mirror,  contains  an  unknown  quantity  of  water ;  from  the  angle 
at  which  a  ray  of  light  must  enter  through  one  of  two  small 
holes  in  the  lid  in  order  that  after  refraction  and  reflexion  It 
may  emerge  at  the  other,  determine  the  height  of  the- water  in 
the  box. 

Let  a  be  the  height  of  the  box,  x  the  depth  of  the  water, 
2h  the  distance  between  the  small  holes  in  the  lid  j  ^,  (f)'  the 
angles  of  Incidence  and  refraction  when  the  ray  enters  the  water. 


1850.J  GEOMETRICAL  Ol'TICS.  349 

they  will  also  be  the  angle  of  refraction  and  incidence  as  it 
emerges  from  it :  hence  we  have 

h  =  [a  —  x)  tan  0  +  ic  tan  0', 

sin0  =  /isin^'; 

a  tan  6  —  h 
tan^  —  tan^' 

a  tan  (j)  —  b 

,  sin  cb        ' 

tan</)-7-^ ^^---1 

(/Lt  —  sni  9)* 

whence  x  is  known  from  <^,  which  is  the  angle  the  ray  makes 
with  the  vertical  upon  entering  the  hole  in  the  lid. 

3.  If  a  kiminous  point  be  reflected  by  a  small  plane  mirror, 
so  as  to  be  seen  by  an  eye  in  a  given  position,  and  the  mirror 
move  in  such  a  way  that  the  Imninous  point  always  appears  to 
be  upon  a  given  conical  surface,  of  which  the  point  is  the  vertex 
and  a  line  through  the  eye  the  axis ;  find  the  form  of  the  sm'face 
upon  which  the  small  mirror  must  always  be  situated. 

Let  0,  E  (fig.  114)  be  the  position  of  the  luminous  point  and 
eye  respectively,  31  any  position  of  the  miiTor,  P  the  corre- 
sponding position  of  the  image  of  0 :  then  will  EMF  be  a 
straight  line,  and  MP  =  MO.  Take  OEx  for  axis  of  x,  Oy 
perpendicular  to  it  for  that  of  y ;  x^  y  the  coordinates  of  M\ 
x\  y'  those  of  P.     Then 

MP  =  MO ; 

.-.  {x'-xy+{y'-yY  =  x'  +  y% 

or  x'  +  y"'  -  2xx'  -  2yy'  =  0 (1): 

JJ,  M^  P,  are  m  the  same  straight  line, 

.-.  -y-^J— (2). 

X  —  a      X  —  a  ^ 

Let  the  equation  to  the  generating  line  of  the  cone  on  which 
P  is  situated,  be 

y  =  mx (3) : 

between  (1),   2),  (3),  we  have  to  eliminate  x',  y. 


350  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1850. 

From  (2)  and  (3), 

x  -  a  _  m[x  —  a)  ^ 
X  y 


y  —  ra[x  —  a)  ' 
From  (1)  and  (3), 

(1  +  Q)f)  x  —  2[x  +  my)  =  0, 

or  2{x  +  my)  [y  —  m[x  —  a)]  —  (1  +  vi^)  a?/  =  0 ; 

.-.  2m{y''  —  x']  +  2{l—  m')  xy  +  2'tnax  —  (1  - m')  ay  =  0, 

the   equation  to  an  equilateral   hyperbola;    and   the   required 
surface  is  an  equilateral  hyperboloid  of  revolution. 

4.  If  the  earth,  supposed  spherical,  were  covered  to  a  depth 
h  with  water,  h  being  small  compared  with  the  earth's  radius, 
shew  that  the  height  to  which  a  person  must  be  raised  above  the 
surface  of  the  water  in  order  to  see  as  far  below  the  horizon  as 

when  he  was  on  the  surface  of  the  earth  is  ^    ,  .. r  nearly, 

2r{fM'-  1)  •^' 

fi  being  the  index  of  refraction  for  water. 

Let  0  (fig.  115)  be  the  centre  of  the  earth,  A  the  station  of 
the  observer  on  the  earth's  surface :  in  order  to  see  as  far  below 
the  horizon  as  possible  he  must  look  in  the  direction  AB,  such 
that  OB  A  is  the  critical  angle ;  he  will  then  see  objects  situated 
in  the  line  5(7,  if  i? (7  be  a  tangent  to  the  surface  of  the  water 
at  B.  Hence  P,  the  raised  position  of  the  observer,  must  be  the 
intersection  of  CB,  OA.     Let  A  OB  =  6 :  then,  if  AP  =  x, 


or,  since  ^,  - ,  - ,  are  small 


r  + 
r  + 

h  _ 

X 

=  cos^. 

lall, 

X  - 

r 

-h 

=1- 

X  - 

-k 

=  ie'. 

1-— -1=1-^6^; 


1851.]  GEOMETRICAL  OPTICS.  351 

Also  from  the  triangle  OAB^ 

■    /  •  -1  1 
,       sin    sni    —  + 
r  +  h  \         /x 


•       •  -il 

sin  sin    - 


(A 
h 


or  l+^=l  +  (/.'^-l)i^, 


or     ^   =     -—a TTT—  3 


or  x  —  Ti  — 


2r'  {fjC'  -  1)  ' 


2r  (/Lt'  -  1) ' 


the  required  distance  to  which  the  man  must  be  raised  above  the 
surface  of  the  water. 

1851. 

A  number  of  vertical  plane  reflectors  are  placed  together  so 
as  to  meet  a  horizontal  plane  in  a  polygon  of  n  sides :  find  the 
path  of  a  ray  of  light  which,  after  reflexion  at  the  n  plane 
reflectors  in  succession,  will  continue  to  proceed  in  its  onginal 
course. 

Shew  also  that  when  there  are  four  reflectors  the  problem 
is  either  indeterminate  or  impossible ;  and  that  when  the  number 
of  reflectors  is  even,  and  the  polygon  capable  of  being  inscribed 
in  a  circle,  the  problem  is  indeteraiinate. 

Let  ^j,  O,y..0^  be  the  complements  of  the  successive  angles 
of  incidence  or  reflexion,  a^,  a,^...a„  the  angles  of  the  polygon: 
then  ^j,  ^.^,  flj,  are  the  angles  of  a  triangle ; 

so    0„  +  0^  =  TT  —  a 


.(A). 


0,   +   ^,   =  TT 


352  SOLUTIONS  OF  SEXATE-IIOUSE   PROBLEMS.  [1851. 

Again,  let  ?/,  =  0,  u^  =  0,...  u^  =  0,  be  the  equations  of  the 
successive  parts  of  the  ray's  path,  a^  =  0,  a^  =  0,...  a^  =  0,  the 
equations  of  the  sides  of  the  polygon ;  the  equations  being  all  in 
such  a  fonn  that  u  considered  as  a  function  of  x  and  y  is  the 
distance  of  the  point  [xy)  from  the  line  u  =  0. 

Now  Oj  is  the  external  bisector  of  w^,  w^,  whence 

Wj  +  u^  =  2a^  cos  0^. 

For  let  Owj,  Oti^^  Oa^^  (fig.  116)  be  the  lines  w.,,  Wv,,  o, ;  take 
any  point  P,  join  OP,  and  draw  P>-,  Ps^  Pt,  perpendicular  to 
these  lines  respectively.     Then 

Pr  +  Pt=  OP  (sill  POr  +  sin  POf) 

=  2  OPsin^iPOr  +  POf)  co^{POr  -  POt) 
=  2  0PsmPOs  costOs 
=  2Pscos6^. 

Hence,  if  Xj  y,  the  coordinates  of  P,  be  substituted  in  Mj,  u^^  a^, 

we  shall  have 

u^  +  u^  =  2a^  cos^j, 

and  the  same  may  be  shewn  wherever  the  point  P  is  taken ; 
hence  generally, 

u^  4-  ti^  =  2a^  cos 0^:'\ 

so  u^  +  u^  =  2a^cose^  \ 
■?/,  4-  u   =  2a  cos^ 

»     '         1  n  n   J 

From  equations  (A)  6^^  ^.2V  ^„  must  be  found,  and  thence 
Wj,  «*2V  ^n  from  equations  (B),  and  the  path  of  the  ray  will  then 
be  fully  determined. 

If  there  be  four  reflectors  we  find,  by  adding  the  P'  and  S'^ 
of  equations  (A), 

O,  +  e^  +  0,  +  0,  =  27r-a^-^,: 
.similarly,  by  adding  the  2"<*  and  4}^^ 

^,  +  ^.  +  ^3  +  ^4  =  2vr  -  a.^  -  a^  : 


1851.]  GEOMETRICAL   OPTICS.  353 

hence  we  must  have 

a,  +  a,  =  a^  -f-  a^  =  TT, 

or  the  quadrilateral  must  be  inscrlbable  in  a  circle  in  order  that 
the  problem  may  be  possible. 

If  however  this  condition  is  satisfied,  still  the  problem  is 
indetemiinatc ;  for  if  wc  treat  equations  (B)  as  above,  we  find 

ttj  cos^j  +  rtgcos^j  =  a^  cos6^  +  a^cos^^ (1). 

This  is  an  identical  equation ;  It  will  therefore  lead  us,  by  means 
of  equating  the  coefficicjits  of  x  aiid  i/  on  the  two  sides  of  the 
equation,  to  three  conditions  between  ^,,  6,^^  6^,  6^^  and  constants : 
between  these  constants  there  ai'e  also  two  relations  arising  from 
the  circumstance  that  the  quadrilateral  may  be  inscribed  in  a 
circle,  and  the  three  conditions  between  ^,,  6^^  6^^  6^^  and 
constants  amount  to  only  one  equation  independent  of  equa- 
tions (A). 

This  condition,  together  with  equations  (A),  will  detci*mine 
^1?  ^2i  ^35  ^4-  ^^^  since  we  have  derived  the  condition  (1)  from 
equation  (B),  it  shews  that  those  equations  are  equivalent  to 
only  three  independent  equations,  and  ii^^  u^,  u^,  u^  are  therefore 
indetemiinatc.  The  dii-ection  only  of  the  different  parts  of  the 
path  of  the  ray  can  be  determined;  with  these  directions  any 
position  will  satisfy  the  problem. 

Similarly,  if  there  be  any  even  number  of  reflectors,  we  may, 
from  equations  (A),  deduce  the  condition 

a,  +  0(3  +  ...  +  a„_,  =  (x^  +  a^  +  ...  +  a„ 

or  the  siuns  of  the  alternate  angles  must  be  equal :  this  condition 
is  satisfied  if  the  polygon  can  be  inscribed  in  a  circle,  and  the 
problem  is  then  possible. 

The  problem  is  still  indeterminate,  for  from  equations  (B) 
we  may  deduce  the  condition 

a,cos^,+a3Cos^34-...+  a„_,cos^,,_,=rtjjCos^2+a^cos^_,  +...+  «„co8^,, : 

this  Identical  equation  will  lead,  as  before,  to  one  equation  of 
condition  independent  of  equations  (A)  between  6^^  0^,  ...  0^^ 
and  constants,  which,  with  equations  (A),  serves  to  detennine 

AA 


354  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

O^^O.^y..  6^.  Still  equations  (B)  are  equivalent  to  only  n  —  \ 
independent  equations,  and  ?ij,  ?<2,  ...  w„  are  therefore  inde- 
terminable. Ill  fact,  we  have  shewn  in  Problem  2,  page  59, 
that  '  if  a  ray  of  light,  after  being  reflected  any  number  of  times 
in  one  plane,  at  any  number  of  plane  surfaces,  return  on  its 
fonner  course,  the  same  will  be  time  of  any  ray  parallel  to  the 
former  which  is  reflected  at  the  same  surfaces  in  the  same 
order,  provided  the  number  of  reflexions  be  even.' 


(     355     ) 


ASTRONOMY. 

1848. 

1.  If  there  had  been  uo  stars,  how  might  the  absoUite 
periodic  times  of  the  Earth  and  planets  have  been  determined, 
even  if  the  eqnator  had  coincided  with  the  ecliptic  ? 

We  might  first  have  detenuined  the  synodic  time  ( T)  of  the 
Earth  and  any  superior  planet  by  observing  the  interval  between 
successive  conjunctions.  Let  jE,  P  be  the  periodic  times  of  the 
Earth  and  the  planet, 

27r       27r  ,     .  ,  ,     .  . 

-'.  'Y  ~  'p  ~  relative  angular  velocity  of 

the  Earth  and  planet 
-    yr, 

or  E=t{i-^ 

We  might  then  have  observed  the  elongation  from  the  Sun 

of  P  and  the  other  planets  at  their  points  of  station :  this  gives 

the  ratio  of  the  distances  of  the  Earth  and  each  planet  from  the 

Sun,   and  therefore,  by  Kepler's  law  that  the  squares   of  the 

periods  are  as  the  cubes  of  the  mean  distances,  it  would  give 

E 
approximately  the  ratio   of  the  periods  or   -p  for  each  of  the 

planets;    whence    from    above   E^   and    the  periods   of  all   the 
planets,  would  be  known. 

2.  A  star  map  is  laid  down  on  the  gnomonic  projection,  the 
plane  of  projection  being  parallel  to  the  equator:  give  a  gra- 
phical  solution  of  the   problem,    to    determine  the  time   at   a 

AA2 


35G  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

kno\NTi  place  l>v  observing  wlicn  two  stars  laid  down  in  the  map 
arc  in  the  same  vertical  plane. 

Since  the  place  is  known,  we  may  draw  about  the  centre  of 
the  map  the  circle  described  by  the  projection  of  the  place  on 
accomit  of  the  Earth's  daily  rotation.  Let  a  line  through  the 
two  stars  intersect  this  circle  in  two  points  A  and  B.  Then  it 
is  plain  from  the  method  of  projection,  that  when  the  projection 
of  the  place  is  at  A  or  B,  the  two  stars  arc  in  the  same 
vertical. 

Let  the  differences  of  longitude  of  the  Sim  in  its  place 
among  the  stars  on  the  day  of  observation,  and  the  points  A 
and  B,  be  obseiwed ;  this  longitude,  converted  into  hours  at  the 
rate  of  15  degrees  to  an  hour,  w^ill  give  the  interval  since  last 
noon  or  the  true  solar  time. 

3.  Shew  that  at  the  equinoxes  the  extremity  of  the  shadow 
of  the  style  of  a  vertical  south  dial  will  trace  upon  the  dial-plate 
a  horizontal  straight  line  at  a  distance  acosec?  from  the  upper 
extremity  of  the  style,  a  being  the  length  of  the  style  and  /  the 
latitude  of  the  place. 

On  the  day  of  the  equinox  the  Sun  appears  to  move  in  a 
great  circle  of  the  celestial  sphere.  We  may  consider  the 
extremity  of  the  style  as  the  centre  of  that  sphere,  or  that  the 
Sun  moves  in  a  plane  through  the  extremity.  The  nonual  to 
this  plane  lies  in  the  vertical  plane  thi'ough  the  north  and  south 
points,  therefore  its  intersection  with  the  dial-plate  will  be  a 
horizontal  straight  line ;  this  is  the  line  traced  out  by  the  ex- 
tremity of  the  shadow  of  the  style. 

Let  the  plane  of  the  paper  be  the  vertical  plane  containing 
the  style  AB  (fig.  117).  Draw  AG  vertical  and  BC  perpen- 
dicular to  AB:  then,  since  AB  is  parallel  to  the  Earth's 
axis,  and  the  Sxm  is  in  the  equator,  C  is  the  extremity  of  the 
shadow  at  noon,  and  ^  0  is  the  distance  of  the  horizontal  line 
from  A  :  it  =  ABcosecACB  =  a  co&ecl. 


ASTRONOMY.  357 

1850. 

1.  If  a  rod  be  fixed  into  a  vertical  wall  which  faces  the 
south  and  the  shadow  of  it  be  cast  upon  the  wall  by  the  Sun, 
find  the  curve  upon  which  the  shadow  of  the  end  of  the  rod  will 
be  situated  every  day  at  mean  noon,  the  Sun  being  supposed  to 
move  imifonnly  in  the  ecliptic  with  his  mean  motion. 

The  mean  Smi  is  situated  on  the  equator  at  the  same  distance 
from  "Y*  as  the  true  Sun ;  it  is  mean  noon  when  this  mean  Sim  is 
due  south. 

Let  S,  S'  (fig.  118)  be  the  true  and  mean  Suns,  then 
nr  >S'  =  "V  >S",  and  if  we  draw  SD  an  arc  of  a  great  circle  perpen- 
dicular to  the  equator,  and  call  'Y'  S,  tp  D,  SD^  i,  a,  S,  re- 
spectively, we  have 

S'I)=yS-'rD  =  L-a. 
Now  let  E  be  the  extremity  of  the  rod,  0  its  shadow  when 
the  Sun  is  in  v ,  P  the  position  of  its  shadow  when  the  Sun  is 
at  yS,  and  S'  on  the  meridian,  i.e.  due  south.  Then  if  we  draw 
ON  horizontal,  NF  vertical  in  the  plane  of  the  wall,  and  join 
EO,  EN,  EP]  OEN==  S'D,  NEP  =  SB,  EPN=l  -  S,  where 
I  =  latitude  of  the  place.     Call  On,  x,  JVP,  y,  EO,  d, 

.'.  X  =  dtoxiS'd, 

^{d'  +  xy-smSD 

,,      ,  -r        .        ,    tanZ  —  tana 

or  X  =  dtaiiiL  —  a)  =  d — — ; — -^ 

^  '  I  +  tana  tsuiL 

,  tana  —  cos w  tana      .  , 

=  d 5 ,  smce  cosw  taniy  =  tana 

cos<w  +  tan  a 

,  (1  —  cosci')  tana  ,  . 

=  a  — f — 5 —  (1) ; 

coscD  +  tan  a  ^  ' ' 

and  3/  =  {d'  +  xr~    -^f.,  =  [cl'  +  ^y   .    J      \ -, 

^       ^  sm(/— o)  '   sm^cotd  — cosZ 

=  [d'^+x;')^  -r-5 — J—. —  ,  since  sina  cotS=cot&>...(2). 

sm/cotw— cos/sma  ^  '' 

"We  have  now  to  eliminate  a  between  (1)  and  (2). 


358  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1850. 

From  (2) 

sin 7  cot G)  ,      (d'^  +  x'^)^ 

— ; cost  =  -^^ , 

sma  y 

•      7  V 

or  sin  a  =  sin  6  cotco  -r^. rrd- 


{(.r  +  xy  +  ycosl' 
From  (1) 

(coso)  cos^rt  +  sin^a)  x  =  d{l  —  cosw)  sina  cosa, 

or  (coso)  +  (1  —  COSO))  sm'aYaf  =  (f  (1  —  cosw)''  sin^a  (1  -  sin^a) ; 

.-.  [cosw  {{d^  +  xy+ycosl]'^  +  (1  —  cosw)  sin'"*?  cot^ co.^^]'"'  x^ 

=  (1  -  COSO))''  cof  0)  sin^?  d'f  [{{d^  +  x^  +  y  cos?]''  -  sm7  cof  eo.?/''], 

the  equation  to  the  required  cm've. 

2.  Suppose  that  dm'ing  the  day  of  the  equinox,  a  man  walks 
in  a  horizontal  plane  towards  the  Sim  at  a  miiform  rate ;  prove 
that  the  equation  of  the  path  described  by  hun  is 

ny  \  _  sin?  ^^  ^^^^^  _^   ^   „,,^, 


sin   -^  +  0=^    £«'-'  +  e  ,, 

Vasecf       /         2     V  / 

where  x  and  y  are  the  coordinates  of  his  position  at  any  time, 
measured  along  and  at  right  angles  to  his  meridian  at  noon ; 
I  is  his  latitude,  and  a  is  the  space  he  walks  over  while  the 
Earth  revolves  through  an  angle  n. 

Deduce  the  particular  cases  of  his  being  at  the  pole  and  at 
the  equator. 

Let  0)  be  the  angular  velocity  of  rotation  of  the  Earth  about 
its  axis ;  the  angle  apparently  described  by  the  Sun  in  the  time 
t  will  be  ft)^,  since  the  Sim  is  in  the  equator,  it  being  the  day 
of  the  equinox. 

If  a  be  the  Sun's  azimuth  at  time  <, 

dy 

-~  =  tana. 

ax 

Now  I  and  mf  are  the  sides  of  a  right-angled  triangle,  sup- 
posing the  man  to  start  at  noon,  of  which  the  angle  opposite 
tot  is  a  y 


1850.]  ASTRONOMY.  359 

t&ncot 
.•.  tana  =  -v— ,-  , 

dy      tanwf 
dx       sin?   ' 

■V  dy  _  tanw^ 

*°^  ds  ~  (sm^Z  +  tau'^a)«)i ' 

cfe  sinZ 


(/5       (sin' ?  + tan' «»«)** 


Also,   *  =  -  G)^ 


n  .    n 

-  tan  -s  sin  -  s 

ay      a a 

^       fsin^Z  +  tan'-sV  fsin'^Z  cos' -  s  +  1  -  cos'' -  sV 

V  a  J  \  a  a  J 

.    n  .    n 

sua  —  s  sm  -  s 

a  ,  a 

—  sec6  4 ; 


[  1  —  cos''?  cos''- s]  f  sec' ?  —  cos' - s j 

.*.  y  = -secZ-jcos'M  cos  -  scosZj  —  4  ;    •.•  y  =  0  when  s  =  0, 

or  cos -s=  sec?  cosf — —i+l] (1). 

a  \asecl      J 

^                       sm  I  cos  -  s 
.      .     dx a 

[sm?  1  —  sm  -s    +  sm  - sj 
V  a  I  a 

sm?  cos-« 
a 


( sin'' ?  +  cos' ?  sin'' -s\ 


n 

cos  -s 

7  ^ 

=  tan? 


[tan' ?  +  sin''  - s  j 


360  SOLUTIONS   OF  SENATE-HOUSE   PROBLEMS.  [1851. 


a 


.    w        /      .         .  .  w   \' 
sin  -  s  +    tan"''  I  +  sin"  -  s  ) 

a        \  a  ) 


.'.  X  =  -  tan  Hog  i  ;  '.'  x  =  Q  when  s  =  0 ; 


n 


tan  / 


/  \4  "' 

.'.  sin  -s  +  (tan'/  +  sin'  -  s]   =  tan?  e°  '^°'  ? 
a         \  a   J 

.-.  8in-5-  (tan'/+sin^-5)   =  -  tanZ  e""'^"' , 

a  \  a   J 

or  (sec'Z-cos'^sV  =  ^  /ga-^  +  g-^  ^ 

the  required  equation  to  the  path. 

We  have  in  this  sokition  considered  I  constant ;  if,  however, 
the  man  be  at  the  pole,  I  will  =  ^tt,  and  sec/,  tan/  will  be 
susceptible  of  great  changes  when  /  alters  but  very  little ;  hence 
we  must   consider  his  motion   as  indefinitely  small   compared 

with  that  of  the  Sun,  or  -  indefinitely  small :  hence  the  above 

equation  leads  us  to  x  =  0,  ^  =  0 ;  and  the  man  merely  stands 
at  the  pole  looking  towards  the  Sun. 

If  he  be  on  the  equator,  tan  /  =  0,  and  therefore  a;  =  0,  or  he 
walks  along  the  equator. 

1851. 

The  declination  of  the  Sun  at  two  obsei'vations  S,  8',  and 
the  Sun's  motion  in  right  ascension  and  longitude  in  the  in- 
tei-val  between  the  observations,  are  equal:  shew  that  if  &)  be 
the  obliquity,  and  a,  /  the  Sun's  right  ascension  and  longitude 
at  the  first  observation,  cosw  =  cosS  cos 8';  tana  =  sinS  cot 8'; 
cot/  =  sin 8'  cotS. 

Let  P,  K  (fig.  119)  be  the  poles  of  the  equator  and  ecliptic; 
>S',  S'  the  two  positions  of  the  Sun :  then,  since  the  differences 


1851.]  ASTRONOMY.  361 

of  the  Sun's  longitude  and  right  ascension  at  S  and  S'  are 
equal,  SS'=SPS'.  Let  the  angle  P8K=^'^  draw  KR^  an 
arc  of  a  great  circle,  to  meet  SP  produced  at  right  angles: 
then,  in  the  triangle  SPS\ 

sin/S-S"  ^mPSS'  =  sin -S'P  sin /SP/S", 

or    cos<^  =  co88';    .*.  ^  =  2', 

and     KR  =  (f)=:B': 

also     PR  =  8,    KP  =  ft), 

lKPR  =  90  -  a,  lPKR  =  I 

Hence,  by  Napier's  rules, 

cos  0)  =  cos  8  cos  S'  (1), 

sin 8  =  tana  tan 8', 

or     tana  =  sin 8  cot 8'  (2),. 

sin 8'  =  cot?  tan 8, 

or      cotZ  =  sin8'  cot8 (3), 

and  (1),  (2),  (3),  are  the  formulae  required  to  be  proved. 


(    362     ) 


DISTURBED  MOTION. 

1848. 

Two  bodies,  P,  P  (fig.  120),  describe  round  a  central  body  S 
circular  orbits  lying  in  one  plane,  the  orbit  of  P  being  within 
that  of  P';  prove  that  the  disturbing  force  of  P'  on  P,  when 
wholly  central  and  additions,  will  be  equal  to  the  disturbing 
force  when  P,  P'  are  on  opposite  sides  of  /S,  provided  SP'  be 
a  mean  proportional  between  SP  and  SP  +  SP. 

Let  Pj  be  the  position  of  P  when  the  distui'bing  force  [F^ 
on  P,  is  wholly  central  and  additious ; 

.    F  -J^    ^ 
•       >      P^P"''P^P"> 

(where  fi  is  the  absolute  force  of  attraction  of  P'). 

Under  the  condition  that  the  force  of  attraction  of  P'  on 
S  and  Pj,  perpendicular  to  SP^,  is  the  same,  i.e.  that  SPP^  is 
an  equilateral  triangle, 

.-.  PP^  =  PS, 

IJ,.SP^ 


or     F  = 


SP" 


Let  F^  be  the  disturbing  force  when  P  is  in  opposition  to  P' 


at  P„ 


P  = 


fi  fi 


if 


SP"      P^P" ' 
and     P,  =  P, 
1  1  ^P 


SP"    {SP^  +  SPY~  SP"' 

or,  dropping  the  suffixes,  because  SP^  =  SP^, 

{SP+  spy  -  SP'  =  ^  {SP+  spy 


1849.]  DISTURBED   MOTION.  363 

or     SRSP'  +  2SF"  =  SF'  +  2SRSP"  +  SF% 
or     SF'=  SF{SP+SF'), 
or  if  SF"  be  a  mean  proportional  between  SF  and  SF  +  SF. 

1849. 

If,  in  addition  to  the  force  of  the  Sun  on  a  planet,  there  be 
a  small  force  tending  towards  the  Smi,  and  varying  invei*sely  as 
the  m^^  power  of  the  distance  of  the  planet  from  the  Smi,  prove 
that  the  perihelion  of  the  orbit  will  have  a  progressive  or  re- 
gressive motion,  according  as  m  is  greater  or  less  than  2. 

Can  you  explain  this  result  by  reasoning  similar  to  that  used 
in  "  Airy's  Gravitation'''''^ 

If  F  be  the  whole  central  force  on  the  planet  we  shall  have 

=  jXU    +  fill    , 

where  fi'  is  very  small.     The  equation  of  motion  is 
d%  F 

d'^u  II       u!    „,_.. 

For  a  first  approximation, 

d^u  yU- 

W  +  "  ~  P  "  ^' 

which  will  be  very  approximately  satisfied  by 

M  =  «{1  +e  cos(c^  — a)], 

if  c  be  very  near  unity,  and  a  =  ^  ; 

.-.  ^^  m"'-»  =  ^  a"'-'  {1  +  {m  -2)  e  cos(c(9  -  a)|, 

omitting  higher  powers  of  e. 

Hence,  for  a  second  approximation, 


-j^  +u-  a-  —  a"'  '  -- 

do  fJL  fl 


+  u-  a-  —  a"'"'  -  —  a"""'  [m  -2)  e  cos  {c0  -  a)  =  0, 


364  SOLUTIONS   OF   SENATE-HOUSE    PROBLEMS.  [1849. 

which  is  satisfied  by 

u  =  a\l+  —  a'"-'  +  e  cos (cd  -  a)]. 

if     ae{l-  c')  cos [cO  -  a)  -  —  a'""'  {m  -2)  e  cos(c^  _  «)  =  0, 

A* 

or     l-c'  =  ^(m-2)a!"-': 

hence  c  is  <  or  >  1  according  as  m  is  >  or  <  2. 

Now,  the  argument  {c6  —  a)  may  be  put  in  the  form 

^-{a+(l-c)^}; 

whence  it  appears  that  the  above  equation  between  u  and  0 
is  the  equation  to  an  ellipse,  the  longitude  of  whose  apse  is 
a  +  (1  —  c)  ^;  its  apse  will  therefore  progress  or  regress  accord- 
ing as  c  is  <  or  >  1,  i.e.  according  as  m  is  >  or  <  2. 

This  result  may  be  explained  in  a  manner  similar  to  that 
used  in  Aiiy's  Chavitation^  Art.  98. 

Let  P,  A  be  perihelion  and  aphelion. 

The  disturbing  force  is  towards  >S'  both  at  P  and  A ;  it  will 
therefore  progress  about  P  and  regress  about  A.  To  consider 
which  of  these  effects  will  be  the  greater.  If  the  disturbing 
force  at  P,  A  and  the  other  points  of  the  orbit  were  propor- 
tional to  the  inverse  square  of  the  distance,  its  only  eflfect  would 
be  to  alter  the  magnitude  of  the  central  force  in  a  certain  ratio 
without  altering  its  law ;  it  would  therefore  have  no  effect  upon 
the  position  of  the  apsides,  or  its  eflfects  about  P  and  A  would 
be  equal.  But  if  the  disturbing  force  vary  inversely  as  the 
(distance)"',  where  m  is  >  2,  the  ratio  of  its  intensity  at  P  to 
its  intensity  at  A  will  be  greater  than  the  ratio  of  the  intensities 
of  the  central  force  at  those  points ;  hence  its  effect  will  be 
greater  at  P  than  at  -4,  or  the  progression  at  P  will  be  greater 
than  the  regression  at  A ;  i.  e.  on  the  whole  the  perihelion  will 
progress.  Similarly,  it  may  be  shewn  that  if  m  be  <  2,  the 
perihelion  will  regress. 


(    365     ) 


ATTRACTIONS. 

1848. 

1.  A  sphere  is  composed  of  an  immense  number  of  free 
particles,  equally  distributed,  which  gi'avitate  to  each  other 
without  interfering:  supposing  the  particles  to  have  no  initial 
velocity,  prove  that  the  mean  density  about  a  given  particle 
will  vaiy  inversely  as  the  cube  of  its  distance  from  the  centre. 

The  attraction  upon  any  particle  will  be  the  same  as  if  the 
matter  nearer  than  itself  to  the  centre  were  collected  there,  and 
attracted  with  a  force  varying  inversely  as  the  square  of  the 
distance.  This  attracting  mass  will  remain  the  same  for  the 
same  particle  throughout  the  motion.  Let  x^  x-\-  ^x  be  the 
distances  from  the  centre  at  the  time  ^,  of  two  particles  situated 
in  the  same  radius,  whose  original  distances  from  the  centre 
were  «,  a  +  S«  ; 

d'^x  fji, 

•'•   ~cie^~  x'' 

and    U-     =2/.    ---     =2^ 


,dtj  \x      aj  ax 

dt  ( ci\^         X 


dx  \2fiJ    (ax—x^)^ 

a  \^  (       ^a  ^a  —  x 


2fjbJ    \{ax  —  xy       {ao;  —  x;y 

But  /A  depends  upon  the  mass  originally  contained  within  the 
sphere  radius  a ; 

.*.  fi^  a^  =  Ca^  suppose ; 


366  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1848. 

In  order  to  find  the  relation  between  ^u-  at  the  time  t  and 
8a,  we  must  differentiate  this  equation,  considering  x  and  a 
variable,  and  t  constant; 

_  /,,  lx\  aSx  —  xBa 
•■■  ^       -^  W  a'         ' 

or     8x  =  '-  8a. 
a 

Hence  the  volume  of  the  shell  originally  contained  between 
the  spheres  of  radii  a,  a  +  Sa,  i.  e.  of  volume  iTrci^Sa^  is  now  of 

x^ 
volume    Attx^Sx  =  47r  —  S«  a  x^.      Hence   the   density   of  the 

matter  in  this  shell,  Avhich  varies  inversely  as  the  voliune,  varies 
inversely  as  its  (radius)' :  hence  the  proposition  is  time. 

2.  Prove  geometrically,  or  otherwise,  that  if  g  be  the  attrac- 
tion which  a  particle  m  exerts  on  a  point  m  a  closed  surface  8, 
0  the  angle  between  the  direction  of  g  and  the  normal,  doi 
an  element  of  8^ 

JJg  cos,  Odd)  =  4:77771,    or  =  0, 

according  as  m  is  within  or  without  8,  the  attraction  of  m  at  the 

ni 
distance  r  being  — ^  . 

Extend  this  result  to  the  case  of  a  finite  mass  cut  by  8,  and 
thence  prove  by  taking  for  8  an  elementary  parallelopiped,  that 
if  V  be  the  potential  of  any  mass  for  an  internal  particle, 
d'V      d'V      d'V 

'd^-^W^w=-^''p' 

About  the  particle  m  as  centre  describe  a  sphere  of  radius 
unity ;  and  let  a  cone  having  m  in  in  its  vertex,  and  circum- 
scribing the  element  dco  of  the  smiace  8,  include  a  portion  dco' 
of  the  surface  of  this  sphere :  then  the  relation  between  day 
and  dw'  will  be 

dot)  cos  6  =  r\  do)'. 


1848.  ATTRACTIONS.  367 

Also,  let  g  be  tlic  attraction  which  m  exerts  on  a  point 
at  distance  unity; 

.■.g  =  l: 
'       r 

hence    g  coaOdo)  =  g'doi'^ 

and    fjg  cosddo)  =  g'co' 

=  the  whole  attraction  of  m  on  w', 

where  w'  is  the  whole  projection  of  S  on  the  surface  of  the 
sphere:  hence,  if  m  be  external  to  S  we  see,  by  taking  the 
projection  of  each  element  with  its  proper  sign,  that  to'  =  0 ; 
but  if  m  be  within  S,  cu'  =  47r ;  and,  by  the  question,  g'  =  m ; 

•'•  !J9  cos^c?&)  =  47rw,     or  =  0, 

according  as  m  is  within  S  or  without  it. 

This  equation  expresses  the  value  of  the  sum  of  the  attrac- 
tions of  a  particle  m  on  the  different  points  of  a  closed  surface, 
each  resolved  in  the  normal  to  the  surface  at  the  point. 

Now,  suppose  the  sm'face  S  to  cut  from  a  finite  mass  the 
mass  M^  the  above  equation  holds  for  every  element  of  this  mass, 
and  therefore  for  the  whole,  if  the  symbols  involved  be  properly 
modified :  we  shall,  therefore,  still  have  the  sum  of  the  attrac- 
tions on  each  point  of  ^S",  resolved  in  the  nonnal  at  that  point, 
=  47rJ/. 

Again,  suppose  S  to  be  an  elementaiy  parallelepiped  so 
small  that  the  density  [p]  may  be  supposed  uniform  throughout 
it:  let  V  be  the  potential  of  a  mass  for  an  internal  particle 
whose  coordinates  are  a?,  y,  z.  Let  P  (fig.  121)  be  the  point 
a;,  ^,  s,  and  the  comer  of  a  parallelopipcd  whose  edges  Zx^  Sy,  S^, 
arc  parallel  to  the  coordinate  axes. 

The  above  considerations  shew  that  the  sum  of  the  attrac- 
tions on  the  faces,  each  resolved  in  a  direction  pei-pendicular  to 
the  face,  will  be  due  to  the  matter  contained  in  the  parallele- 
piped: now 

dV        rrr        p[x -  ^)  d^  dy  d^ 


-III 


dx-       JJj{[j--^f+{^-r,Y+{z-^y]i^ 


368  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [18-48. 

the  integration  extending  throughout  the  parallelepiped ;  hence 

dV     . 
at  P,  -J—  will  be  positive,  since  ^  is  always  greater  than  x ;  at 

P  it  will  be  negative,  since  ^  is  always  less  than  x  +  8x. 
Hence  the  absolute  magnitude  of  the  attractions  parallel  to  the 
axis  of  a^  at  P  and  j)  will  be,  respectively, 

dV      ^       dV      d'V^ 

-J—  and J =-T-  ex. 

ax  ax       ax 

Now  we  may  consider  the  surfaces  il/P,  mj)  so  small,  that 
the  attraction  on  every  point  of  each  of  them  is  the  same :  hence 
the  whole  attractions  on  JWP,  mp  parallel  to  the  axis  of  x 

d^V 
=  -—hxhj  8z. 

The  whole  expression  for  JJg  cos  Odo)  is  in  this  way  found 
to  be 

/     d'V     d'V     d'V\  .    .    ^ 

[-d^-df-^)^''^^  ^^'  ^^"^^  •••  =  ^'^^^ 

=  4:7rp  8x  By  Bz ; 
d'V     d'V     d'V 
•'-   d^-^df-^d^  =  -  ^^^- 

3.    Supposing  a  mass  of  homogeneous  fluid,  which  attracts 

every  particle  of  matter  with  a  force  varying  as     -..     ..;,  to  be 

enclosed  within  a  thin  spherical  shell,  find  the  path  described  by 
a  heavy  body  let  fall  from  any  point  of  the  surface  of  the  fluid, 
the  resistance  varying  as  the  velocity.  Prove  also  that  the  body 
will  reach  the  axis  and  equator  of  the  spheroid  after  the  same 
intei-vals  respectively,  from  whatever  points  of  the  surface  it 
begins  to  fall. 

The  attractions  of  the  spheroid  on  any  particle  within  it 
perpendicular  to  the  axis  and  equator,  vary  respectively  as  the 
distances  {x,  y)  of  the  particle  from  that  line  and  plane,  =  ixx^  fix 
suppose. 


1848.]  ATTKACTIONS.  369 

Also,  by  the  question,  the  resistance  ou  the  particle  in  motion 
=  kv  =  k  —  :  hence  the  resolved  parts  of  it  are 

J  ds    dx      -  ds    dti  ,  dx      ,  dii 

dt    ds  '       dt    ds  ^  dt^       dt 

Hence  the  equations  of  motion  are 

d  X         dx 

^  +  k^  +  ;.x  =  0, 

d'?/       ,  dy 

Let  a,  yS  be  the  roots  of  the  equation 

z^  +  kz  +  fi  =  0, 
and  a',  /9'  those  of 

z""  +  kz  +  yu,'  =  0 : 

the  above  equations  give 

X  =  Ae""  +  Bt^', 

y  =  A's*'  +  B'zl^'\ 

A,  B,  A\  B\  being  arbitrary  constants  to  be  determined  by  the 

circumstance  that  the  body  falls  from  rest  from  a  given  position. 

The  circumstance  that  it  falls  from  rest  gives  us  the  condition 

that  ^  =  0,  ^  =  0,  when  t=0: 
dt        ^   dt        ^  ' 

.'.  0  =  AoL  +  ^/3, 

and    0  =  A' a  +  B'^' ; 

A  B      ^ 

.-.  -  =  -  -  =  C  suppose, 

A'  B'       „, 

W  ^~V^        suppose ; 

the  equations  for  the  determination   of  the  relation   between 

X  and  y  by  the  elimination  of  t. 

Hence,  if  f,  t'  be  times  of  fallmg  to  the  axis  and  equator 

respectively, 

0  =  fit"  -  ai^\ 

BB 


370  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1849. 

.-.  log/3  -\-  at  =  loga  +  yS^; 
log  a  —  logyS 


t  = 


so    «  = 


a-y8        ' 
loga'  —  logyS'  _ 


x'  -  ^'        • 

whence  it  appears  that  f,  t'  arc  independent  of  the  particle's 
original  position. 

1849. 

1.  Each  particle  of  two  indefinite  straight  lines,  lying  in  the 
same  plane,  attracts  with  a  force  which  varies  inversely  as  the 
distance.  Determine  the  motion  of  a  body  projected  in  any 
direction  along  the  plane. 

We  must  first  find  the  attraction  of  either  of  the  lines  AB 
(fig.  122)  upon  the  particle  in  any  position  P.  From  P  draw 
PD,  the  perpendicular  on  AB^  and  join  PQ^  Q  being  a  point 
at  the  distance  x  from  D.  Let  hA  be  the  attraction  of  an 
element  hx  of  the  line  about  Q  resolved  in  PD: 

.:  BA^^cobQPB 

fjbadx 

.'.    A  =  u  tan~^  - 
a 


fJiTT. 


from    a;  =  —  00 
to        a;  =  +  CO 

Hence  P  will  be  attracted  by  two  constant  attractions  in 
constant  directions,  which  are  therefore  equivalent  to  a  constant 
attraction  in  a  constant  direction,  viz.  2yu,7r  sin  a  (2a  the  angle 
between  the  lines),  parallel  to  the  internal  bisector  of  the  lines. 
Hence  the  case  is  the  common  case  of  projectiles,  and  the  path 
will  be  parabolic. 

2.  The  attraction  of  a  uniform  filament  of  matter,  in  the 
form  of  a  plane  curve,  upon  a  particle  is  replaced  by  that  of 
a  circular  filament  having  the  particle  for  its  centre:  find  the 


1850.]  ATTRACTIONS.  371 

law  of  density  of  the  circular  filament  in  order  that  this  may 
be  done. 

Let  the  cuitc  be  referred  to  the  particle  as  pole,  and  let  a 
be  the  radius  of  the  circle,  yu.  the  density  of  the  filament  In  the 
form  of  the  cun^e,  p  that  of  the  circular  filament  at  the  point  [B] ; 

pa  80      fJ'^s  ^ 


■'       ^^           ,:^    > 

fia  ds 

r 

=  fia  \n^  + 

[dej  1 ' 

u 

1 

7- 

? 

fia 

^J' 

if  p  be  the 

perpendicular  from  the 

!  particle 

on 

the 

tangent 

at  the 

point  (r. 

,0) 

5 

1 

1850. 

A  uniform  rod  is  placed  witli  its  middle  point  against  a  rough 
circle,  in  whose  centre  resides  a  force  attracting  inversely  as  the 
square  of  the  distance :  if  the  rod  be  slightly  disturbed  from  the 
position  of  equilibrium,  find  the  time  of  a  small  oscillation. 

Let  6  be  the  inclination  of  the  rod  to  the  horizon  at  the 
time  t]  ^  the  distance  of  its  middle  point  from  the  point  of 
contact  with  the  circle.  Since  the  motion  is  small,  we  may  take 
the  equations  of  motion  about  the  instantaneous  axis  of  rotation : 
hence  we  have 

where  k  is  the  radius  of  gyration  of  the  rod  about  its  middle 
point,  and  L  the  moment  of  the  attractions  on  the  rod  about 
the  point  of  contact. 

bb2 


372  SOLUTIONS  OF  SENATE-HOUSE   I'KOBLEMS.  [I8f)l. 

To  find  L.     In  fig.  120  the  moment  about  D  of  the  attrac- 
tion on  an  element  at  Q  (P  being  the  centre  of  the  circle), 

if  /i  be  the  absolute  force  of  the  attraction,  p  the  mass  of  a  unit 

of  length  of  the  rod, 

fipa  X  Sx 

{a'  +  xj' 
_       ,     ^M- ,  2?  being  the  length  of  the  rod, 


{(i'^rf  \^  '   a^  +  r       V        d'^rj 
omitting  p  and  higher  powers  of  |^, 

2fipal    g  _    2/jbpdH    „ 

to  the  same   degree   of  approximation :    and   the   equation   of 
motion  becomes 

"^^     {d'+ry- 

or,  since  M=  2?p, 

g+        ^^        0  =  0. 

'^^     id'-^rfu' 

Hence  the  time  of  a  small  oscillation 


r=2 


/Lt^a 


1851. 

1.    Two  uniform  straight  rods  AB^  CD  (fig.  123),  mutually- 
attracting  each  other  with  forces  varying  as  the  distance,  are 


1851.]  ATTRACTIONS.  373 

constrained  to  move  in  two  grooves  ABO,  CDO  at  right  angles 
to  each  other ;  dctenuine  the  time  at  which  the  extremity  of  one 
of  the  rods  reaches  0  the  point  of  intersection  of  the  grooves. 

The  attraction  towards  0  of  each  element  p'hri  of  CD  upon 
any  element  ph^  of  AB,  at  a  distance  f  from  0,  will  be  the 
same,  viz.  fip'SrjpB^.^ :  hence,  if  AB  =  2a,  CB  =  2b,  the  whole 
attraction  towards  0  of  CB  on  AB  will 


r 
=  2fipp'b 

J  -, 


^dl. 


if  a?  be  the  distance  from  0  of  the  middle  point  of  AB^ 

=  Afipp'hax. 

The  equation  of  motion  of  AB  is  therefore 

d'^x 
2pa  -y^=  -  ifipp'baxj 

or     -YY  +  w"'a:  =  0,    if  n^  =  2fip'b ; 

.'.  x  =  A  cos[nt  +  B) 
=  x^  cosntj 

if  t  =  0   at  the  beginning   of  motion,    and  x^  is   the   original 
value  of  X. 

Hence,  if  t  be  the  interval  before  A  arrives  at  0,  we  have 

a  =  Xq  cos  nt, 

1  -I  a 

or     t  =  ,- — rfTi  cos    —  . 
{2fip'b)i  x^ 

2.  If  a  portion  of  a  thin  spherical  shell,  whose  projections 
upon  the  three  coordinate  planes  through  the  centre  are  A,  J5,  C\ 
attract  a  particle  at  the  centre  with  a  force  varying  as  any 
function  of  the  distance,  shew  that  the  particle  will  begin  to 
move  in  the  direction  of  a  straight  line  whose  equatioas  are 
X  y  z 
A^B^C' 

Let  0  be  the  angle  which  the  radius  drawn  to  the  element 
SS  of  the  shell  makes  with  the  axis  of  .r;  then,  if  r  =  radius 


374  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1851. 

of  sphere,   and   </>(/•)   be   the   law   of  attraction,  the   attraction 
of  8S  on  the  particle  parallel  to  the  axis  of  x  will  be 

<j){r)  8S  cosO^ 

and  the  whole  attraction  on  it  (X)  parallel  to  the  axis  of  Xj 

X  =  <f){r)  S.8/S'cos0,     since  r  is  constant, 

=  <f>{r)A. 

So      Y=<}>{r)B, 

Z=<f>{r)C. 

And  the  equations  to  the  direction  of  the  resultant  attraction, 
which  is  the  direction  in  which  the  particle  will  begin  to  move, 

are 

X        y        z 

X^  Y^Z' 

X        y        z 


(    375     ) 


PHYSICAL  OPTICAS. 

1848. 

A  spherical  wave  of  light  is  incident  directly  on  a  lens :  find 

approximately   the  retardation  of  the  several  portions  of  the 

•      ,.             ,                             .111 
wave,  and  prove  m  this  way  the  common  equation ~  /  * 

Suppose  the  lens  to  be  a  positive  concavo-convex  whose 
thickness  at  the  middle  point  is  indefinitely  small :  take  this 
middle  point  as  origin  of  coordinates  and  the  axis  of  the  lens 
for  axis  of  x.  The  retardation  of  any  part  or  ray  of  the  wave 
will  =  (/u.  —  1)  X  length  of  the  path  in  glass  =  (/a  —  1)  p  suppose. 

Let  ic,  y  be  the  coordinates  of  the  point  of  incidence  of  this 
ray,  Q  the  inclination  to  the  axis  of  the  part  of  the  ray  within 
the  lens. 

The  equation  to  the  two  surfaces  will  be 

V'  =  ^r^ (1), 

and     77' =  2s| (2), 

very  nearly ;  since  |  is  very  small  for  all  rays  near  the  axis. 

Now  (2)  is  satisfied  by  the  coordinates  x  —  p  cos 9^  y-  p  sin 0, 
or,  as  B  is  veiy  small  as  well  as  p^  hj  x  —  p  and  y ; 

.-.  f  =  2s[x-p)  ='-y-2sp', 

1    ny 


p  =  \r--sn 


therefore  the  retardation  of  this  portion  of  the  wave 

=  (-')(7.-^)f- 

Hence,  if  w  be  the  distance  from  the  origin  of  the  centre 
of  the  incident  wave,   the  equivalent  length  in  air  of  the  ray 


37(>  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1848. 

we  are  considering  from  the  centre  of  the  wave  to  the  point 
of  emergence, 

=  {(«-^)'+/]-  +  /*P 

=  u  —  X  ■\-  ^  +At \  —  very  nearly 

2m  \r       h)   2         "^ 

Consequently  this  ray  upon  emergence  is  in  the  same  phase  as 
the   ray    incident   directly   when    it   has   travelled   a   distance 

\ V  ii\ V—  alter  emergence. 

\u      r      ^\r      s)\  2  ^ 

Now  the  geometrical  focus  upon  emergence  is  the  centre  of 
curvature  at  the  vertex  of  the  surface  of  revolution,  which  is  the 
locus  of  all  parts  of  the  wave  which,  after  transmission,  are  in 
the  same  phase  of  vibration. 

Let  V  be  the  radius  of  this  sphere  when  only  the  parts  of 
the  wave  indefinitely  near  the  axis  have  emerged :  the  sphere 
will  then  pass  through  the  points 

[x-p,  y)  or  (^1^,  3/j  and 

y  being  indefinitely  small ; 


u       r      ^\r       syi    2  ' 


.-.  /  =  2v 


11  ^1  _  l\\  t  4.  t 


u       r^  *^\r       s/j    2   "*"  25 


11/  -N    /^l  1 

/.  -  =  --f  0^-1 

V       u  \r       s 

111  1/11 

or =  -^  ,     if  -,  ^   /A  -  1) 

V      u       J  J  \r       s 

But  evidently  v  is  the  distance  from  the  lens  of  the  geome- 
trical focus  upon  emergence  ;  hence  this  is  the  usual  formvila. 


1850.]  PHYSICAL  OPTICS.  377 

1850. 

A  and  B  being  two  fixed  points,  and  P  such  that  AF=  /m.BP, 
the  locus  of  P  is  a  circle.  Shew  from  this  property  how  to 
construct  a  lens  of  common  glass,  such  that  a  direct  pencil 
incident  from  a  determinate  point  will  be  refracted  without 
aben'ation. 

The  property  enunciated  will  be  found  In  Prob.  8,  p.  157. 

Let  A,  B  (fig.  124)  be  the  points  from  which  the  pencil  is 
to  diverge  before  and  after  passing  through  the  lens  without 
aberration.  Draw  the  circular  arc  HCH'  such,  that  If  Q  be 
any  point  In  It,  BQ  =  fi.AQ. 

With  centre  A  describe  any  circular  arc  HcH'  intersecting 
HCH'  in  H^  H':  HCH'c  Is  the  section  by  the  plane  of  the 
paper  of  such  a  lens  as  Is  required.  For  a  ray  incident  upon 
the  lens  from  A  will  sufiier  no  deviation ;  and 

BQ-{fi.QP+AP)  =fj,.AQ-{fi.QP+AP) 
=  {fi-  l)AP  Is  constant: 
and  therefore,  by  reasoning  similar  to  that  in  Airy's  Tracts^ 
p.  276,  it  appears  that  the  pencil  diverging  from  A  will,  after 
emergence,  diverge  from  B. 

1851. 

If  [6)  be  the  angle  which  one  of  the  planes  of  polarization 
makes  with  the  plane  passing  through  the  normal  to  the  front  of 
the  \f ave  and  either  optic  axis  of  a  blaxal  crystal,  and  y,,  v^  be 
the  two  velocities  of  transmission  of  the  wave,  shew  that 

(y,cos^)'''+  {v^%meY  =  h\ 

Since  the  planes  of  polarization  respectively  bisect  the  acute 
and  obtuse  angles  between  the  two  planes  through  the  normal  to 
the  front  and  the  optic  axes  (Griffin's  Double  Refraction^  Art.  21, 
p.  12),  It  follows  that  the  angle  between  these  two  planes  =  2$. 

Now,  In  accordance  with  the  usual  notation,  the  equations  to 
normal  to  the  plane  front  are 

?  =  J^  =  ? (1). 

I       m      n 


378  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS.  [1851. 

Those  to  the  optic  axes  are 

3^  =  ^^     (;^^4±(J^  =  0 (2). 

If  therefore  tlie  equations  to  the  above  planes  are 

Ax  +  Bi/  +  Cz  =  0, 

A'x  +  i?'?/  +  C'z  =  0, 
we  must  have 

Al  +  Bm  +  Cn  =  0, 

A{d'-b-y  +  C{U'-c')^-  =  0; 
A  B C        _ 

Similarly 

A ^b; C       _  , 

AA'  +  BB'  +  CC 


Also     cos2^  = 


{A  +  B' + cj  {A" + B" + cy- 

AA'  +  BB'+CC 


~  {{AA'+BB'+CC'y+[BC'-CB'y+{CA'-AC'y+{AB'-BA'YY- ' 

Now 
AA'+BB'  +  GC'=  [m'  {¥-  c')  +  r  {W-  c^yd'  {a'-  b')-m'  {d'-b')]  rr' 

=  {[r-^m'){b'-c')  -  {m'  +  n'){d'-¥)]rr' 

=  {{r+m')  b'-{l-n')6'-  (I-  P)  d'+  {m'+n')  &'}  rr' 
'.'  P  +  m'  +  n'  =  1 

=  _  {a'  -b'  +  &  -  [I'd'  +  m'V'  +  d'c^) ]  rr' 

=  -  ( U—  25^)  rr  suppose, 
[BC'-CB'Y  +  {CA'-AC'f  +  [AB'-BA'Y 

=  4  {id'r'  4  m'  +  m'd')  {d'  -  h')[h'  -  c')  rV 

=  Anf{d'-b'){b'-c')rV' 

=  4  Vr\"'  suppose ; 


1851.] 

PHYSICAL  OPTICS. 

co-e         '^'  -  ^ 

^«--^-{(2i^_?7r  +  4F}* 

2V  -  U 

379 


=  (1  _  ^i^  _  ,„'-^)  7/c'  +  .ti^c'a'  +  {l-r-  m']  d'h' 
=  W  suppose ; 

Now  Vj"'*,  vj''  are  the  roots  of  the  equation 

considered  as  a  quadratic  in  v^ ; 

.-.  v^Jrv'^  =  f  [h'  +  c'O  +  m'^  (c^  +  «•-=)  +  n''  {d'  +  J'^) 

=  (l_,,i-''_n'-')(J'''+c2)+(l_^--^_f-')(c''+a^)+(l_Z--'-7;i  ){d'+l'') 
=  a'^  +  ?>■■'  +  c'  -  {rd'  +  m^V'  +  ?i^c''') 

and  y>;  =  T^&^c''  +  y«'''c'^«'''  +  d'd'W 

.  .     cos2c/  = 


v;'  (1  +  cos2^)  +  v^'  (1  -  cos  2^)  =  2l/\ 
and     (y^  cos^)'  +  ((',,  sin  6)'  =  6"^ 


(     380     ) 


CALCULUS   OF   YAIUATIONS. 

1848. 

A  aud  B  are  two  given  points  in  the  generating  line  of 
a  surface  of  revolution  whose  axis  is  vertical :  supposing  a  body 
acted  on  by  gravity,  to  descend  along  the  surface  from  A  to  Bj 
find  its  form  when  the  whole  pressure  upon  it  between  the  two 
given  points  is  the  least  possible. 

Find  also  the  form  of  the  surface  when  the  length  of  the 
generating  line  between  the  point  A  and  B  is  also  given,  and 
point  out  the  difference  between  the  two  results. 

Let  y  be  the  depth  of  any  point  of  AB  below  A,  x  its  dis- 
tance from  the  axis ;  the  pressure  at  this  point  will  be 

P^Mg'^  +  M-, 
as  p 

V  being  the  velocity,  and  p  the  radius  of  curvature  at  the  point : 
hence  we  must  have 

[Pds  =  Mg  I  f  1  +  —  .  -^  I  dx^  a  minimum. 

TT  Tr      ,       2?/  ds 

Here     F  =  1  +  —  -^ 

p    dx 

_i  _  J£^. 

dV  _         1q 


i^  1 


dy  1  +^ 

dp       (1+/)'^' 

^        dq  1  +2^ 


1848.]  CALCULUS   OF   VARIATIONS.  381 

the  equation  between  iV,  P,  Q^  is 

^^     dP     d'Q      ^ 

or     N--Up-^4)=0. 
ax  \        ax) 

dQ  _         2p  ^py 


But     ^-^  =  -,-f^^  + 


dx  1+/   '    (!+/)•'' 


or     <?  =  0, 

shewing  that  the  Une  required  is  the  straight  line  joinmg  the 
points  A^  B. 

K  the  length  of  the  line  be  given,  we  have 

a  being  an  arbitrary  constant  to  be  determined. 

N  and  Q  remain  of  the  same  value  as  before ;  P  becomes 


Hence  we  shall  find,  as  before,  that  <?  =  0,  or  the  required 
curve  is  a  straight  line :  in  this  case,  however,  it  must  be  a 
broken  line,  the  different  parts  of  which  are  equally  inclined  to 
the  vertical,  and  the  inclination  so  chosen  as  to  give  the  line  of 
the  required  length.  The  particle  is  of  course  supposed  to  turn 
the  abiTipt  angles  of  the  line  without  impulsive  pressure  or 
change  of  velocity. 


382  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS.  [1851. 

1851. 

A  uniform  straight  rod  AB  is,  constrained  to  move  in  a 
vertical  plane  with  its  middle  point  in  a  horizontal  groove,  and 
its  upper  extremity  against  a  smooth  curve :  find  the  nature  of 
the  curve  when  the  rod  descends  from  one  given  position  to 
another  in  the  least  time  possible,  the  initial  angular  velocity 
being  given. 

Let  CB  (fig.  125)  be  the  required  curve,  OQ  the  horizontal 
groove  ;  take  the  point  0  in  it  for  origin  of  coordinates :  at  time  t 
let  AB  be  the  position  of  the  rod,  draw  BT  the  tangent  at  P; 
OX  =  X,  XB  =  2/,   0Q  =  ^^  lOQB=  e,  QB  =  a. 

The  equation  of  vis  viva  is 

-j-j   +  ^  \~ji)   =  constant  =  c  suppose (1). 


Also  the  motion  of  B  pei'pendicular  to  BT'is  zero, 

.-.  ^  miBTN  +a~coB QBT  =  0, 
dt  dt        ^  ' 

or,  if    BTN=<j>,     ^sin0  +  a'^^cos(<^-6')  =  O (2); 

also     {^  +  xY  +  /  =  d' (3). 

From  (1)  and  (2) 


\  '^d'co%\<^-e)]  \dt 

or,  since    sin^  =  -,     tanrf>  =  ^  =  w, 
a  dx     ^^ 


X-W 


Also,  from  (3) 

"   dx^  '     {d'-Tf-y 


1851.] 


CALCIJLUS   OF    VAKIATIONS. 


383 


dt 


1  + 


^•y 


[[d^-yy+pyY\ 


d^ 


1 


c[a-y 


^a{(«^-j/?+P3/F+^4>?^^^. 


or 


Here  V  contains  only  y  and  p^ 

.-.     V-Pp  =  C, 

[{(«  -y  npy]  +f^p^-  [{{a^-yy+pyY^jc^^i  +  -  («  -3^ )  -o, 

{a^-yy{{^-yy+py]  +  9.  ^.f)i  [[(^a^^,jy-+pyr+kYY-  =  o ; 


1- 


or  {C'k-y)p  =  {d^-y%     C  =  |l  -  (^)y'; 

.-.     C'k  sm"  '^  +  [p"  -  ff  =  x^  0", 

the  required  equation  to  the  curve :  the  constants  C",  C"  are  to 
be  detemilned  by  the  two  given  positions  of  the  rod  which  give 
two  points  through  Avhich  the  curve  must  pass.  The  curve  is 
independent  of  the  angular  velocity  of  projection. 


(     384 


APPENDIX. 


The  following  problem  in  Astronomy  was  set  in  1848. 

If  a  rectangular  court  be  enclosed  within  a  wall  of  given 
height,  and  one  of  its  sides  be  inclined  at  an  angle  of  30°  to  the 
meridian,  detennine  the  breadths  of  the  shadows  of  the  walls  on 
a  given  day  at  noon,  and  the  portions  of  the  courts  and  walls 
which  will  be  enveloped  in  the  shadow,  the  latitude  being 
52°  30'  north,  and  the  Sun's  declination  on  the  given  day 
7°  30'  north. 

By  referring  to  the  problem  on  p.  64,  we  see  that  here 
0  =  30°  and  </>  =  latitude  —  Sun's  declination  =  45°, 

r.  a  =  \h^    ^  "  2"  ^'' 

Let  ?j,  ?2  ^6  the  lengths  of  the  walls,  whose  shadows  are 
respectively  of  the  breadth  «,  5,  the  area  of  the  courts  enveloped 
in  shade  will  be  \a  +  [l^  —  a)  Z>,  or  l^a  +  I  J)  —  ah ;  and  the 
shaded  parts  of  the  walls  the  whole  of  the  two  walls,  and  two 
triangles  \ha^  \hh  of  the  other  two. 

The  following  solution  of  the  problem  on  p,  148,  is  due  to 
Mr.  Gaskin. 

Let  TP,  TQ  (fig.  126)  be  the  two  given  tangents,  take  the 
line  AB  as  axis  of  cc,  and  let  OP'  Q  be  the  chord  of  contact  of 
any  conic  touching  TP,  TQ^  and  passing  through  A^  B.     Take 


APPENDIX,  386 


0  as  origin,  and  let 


'7     —        )  ''7'    — 

a      0  a       0 

be  the  equations  to  TP,  TQ  respectively  ;  also  let  OA  =  a, 
OB  =  /3.  Let  the  equation  to  OP'  Q'  he  7/  =  mx,  then  that  to 
the  conic  will  be 

,  Hence,  putting  ?/  =  0,  we  get 

o,.  1,    /i  +  i)l  +  (_L,_'|!)  =  o (.), 

a;       V«      aj  X      \aa        \J  ^  '^ 

the  roots  of  which  equation  are  -  ,    p  5  whence  we  see  that 

1111 

a      p      a       a 

or  the  line  OPQ  is  divided  harmonically  in  -4,  i?,  whence  0  is 
one  of  the  foci  of  involution  of  the  system  of  points  P,  Q^  A^  P, 
so  that  the  chords  of  contact  of  all  conies  touching  PP,  TQ  and 
passing  tlu'ough  -4,  P,  cut  AB  in  one  of  the  points  0,  0',  if  0' 
be  the  other  focus  of  involution. 

Now,  in  order  that  (1)  may  represent  a  rectangular  hyperbola, 
the  sum  of  the  coefficients  of  x''  and  y''  must  =  0 ;  hence 

-—  -  m^  +  777  -    1=0. 
an  00 

But  by  (2), 

1    _    1         m' 

a/3       aa         \ 
Combining  these  equations,  we  get 

aa       00  J  \aa       ap 

giving  two  values  for  «i,  equal  and  of  opposite  signs^  so  that 
there   can  be  constructed   two  pair  of  rectangular  hyperbolae 

CC 


386  SOLUTIONS  OF   SENATE-HOUSE   PROBLEMS. 

whose  chords  of  contact  meet  in  one  of  the  foci  of  invohition, 
and  are  equally  inclined  to  AB. 

The  relation  between  the  four  lines  Ox,  Oy,  Op,  Oq,  in  the 
problem  on  p.  158,  may  be  expressed  thus:  Ox  and  Oy  each 
bisect  the  lines  between  0}),  Oq  parallel  to  the  other. 

For  let  the  equations  to  Op,  Oq,  refen-ed  to  Ox,  Oy  as  axes, 
be  2/  =  mx,  y  =  m'x.  Then  if  Oa  =  a,  Ob  =  b,  Od  =  a',  Ob'  =  b', 
the  equations  to  ah,  db'  are 

-  +1  =1 
a^b  ' 


X       y 
a        b 


'.+'Tr=l; 


whence,  if  (xy)  be  the  point  Q  of  intersection  of  these  lines, 
\a      a  J 


But  Q  lies  on  the  line  Oq,  or  y  =  mx, 

^b       b'^ 


■''     («-^'Ka"a')='^^^"^'K^-T')' 


11         /I      lA     ^ 

a       a  \b       bj 


The  condition  that  the  point  of  intersection  of  this  line  lies 
on  Op  or  y  =  mx,  is  derived  from  this  equation  by  interchanging 
a,  a!,  and  writing  m  for  m, 


1       1  , 

?  +  m 

a      a 


■(M)-- 


Hence  m  =  —  m,  which  expresses  the  above  relation  between 
Ox,  Oy,  Op,  Oq. 

The  same  thing  may  be  proved  geometrically  by  making 
any  one  of  the  points  a,  b,  a,  or  b',  remove  to  an  infinite 
distance. 


APPENDIX.  387 

The  following  statical  Problems  set  in  1850  have  been 
omitted. 

1.  A  heavy  rod,  whose  weight  is  TV,  rests  upon  a  fulcrum 
at  its  middle  point,  when  loaded  at  one  end  with  a  weight  W, 
the  density  at  any  point  of  the  rod  at  the  distance  x  from  a 

TTtJC 

certain  point  in  it  varies  as  sin  — ,  a  being  the  length  of  the 

rod :  find  the  ratio  of  W  to  TF',  and  determine  at  which  point 
the  density  is  zero  when  this  ratio  is  the  greatest  possible. 

Let  c  be  the  distance  from  the  centre  of  the  rod  of  the  point 
where  the  density  is  zero,  p  the  density  at  the  point  x  =  ^a. 
The  conditions  of  the  problem  give 

r^      .    TTic  ,         r*-"-'      .    irx  , 
p  sm  —  ax  ■\-  [      p  sm  —  dx  =  W, 
^  Jo  ^ 

pa   (^  (tt      7rc\      ,  /tt      ttcX)        „_ 

or^|l-eos(-  +  -)  +  l-cos(j--)|  =  P^'; 

w    W 
•••     P'^-^ (•)• 

Also  taking  moments  about  the  fulcrum  which  is  at  the 
middle  point  of  the  rod, 

r^"''        .  .  irx  .         r  .       ,  .  TTcc 

Pi      (ic  —  c)  sm  —  ax  =  p  I    [c-  x)  sm  — 
Jc  ti  '^J^^'a 


TTX    , 

ax 


+  p  \       (c  +  ic)  sm  —  ax  -\-  W  - 

/•*'*'      .    irx  J  (  [*-"''  .    irx  J     ^    fi"-'  .    irx  J  \      ^..,  a 

or  pi      xsm  —  ax  —  pel  I      sm  —  ax  +  I      sm  —  ax  ]  =  W  - 

T-T-  f       .     TTX    ,  ax  TTX         a^      .      TTX 

JN ow     ixsm  —  ax  = cos 1 — 5  sm \-  c: 

J  a  TT  a        TT  a  ^ 

.    TTX  ,        a    .    ire  (a  (a 

icsm  —  rta;  =  -sm  —  \-z-\-c-\-\-  -c 
a  IT         «      2  V2 


a       .      TTC 

=  —  sm  — 
TT         a 


888  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS. 

and    I      sin  —  ax  +  l       sin  —  ax  =  —  ,  as  shewn  in  (1 
a''    .    TTC      2  oca      ,,,,  a 


r         /IN      TT  W  fa'   .    TTC      2ac\        „,  a 

or  from  (1        -  —    -  sin \  =  W  -; 

'       2    a  Vtt         a         tt  J  2 


a  sin 2c 

a 


This  ratio  =  oo  when  c  =  0 ;  it  has  a  maximum  value  when 


.      TTC  . 

a  sm 2c  is  a  mimmum ; 

a 


or,  clIfFerentlating  with  respect  to  c, 


TTC        ^        ^ 

TT  COS 2  =  0, 

a 


ire 
COS  —  = 


a       TT 
which  determines  the  value  of  c. 

2.  Portions  are  cut  from  an  ellipsoid  by  planes  which  are 
parallel  and  equidistant  from  the  centre ;  if  -ct  be  the  length 
of  a  pei-pendicular  from  the  centre  upon  either  plane,  and 
Z,  m,  ?i,  the  cosines  of  the  angles  which  it  makes  with  the  axes, 
shew  that  the  remainder  will  rest  when  placed  with  a  section 
on  a  horizontal  plane,  if 

1  r      m'      n' 

—  =  or  >  -^  +  -p-  +  ^ , 
w  a        o        c 

a,  &,  c,  being  the  axes  of  the  ellipsoid;  and  express  the  con- 
dition that  ]j  such  solids,  when  placed  on  each  other  with  their 
sections  coincident,  and  their  centres  in  a  line  inclined  to  the 
vertical,  shall  not  fall  over. 

The  one  portion  will  rest  with  a  section  upon  a  horizontal 
plane  if  the  vertical  line  drawn  through  its  centre  of  gravity, 


APPENDIX.  389 

which  is  the  centre  of  the  ellipsoid,  fall  within  the  section; 
i.e.  if  zj  be  equal  or  less  than  the  radius  vector  (r)  of  the 
ellipsoid  drawn  in  the  same  direction,  or  since 

if  -^  =  or  >  -^  =  or  >  -,  +  77  +  -^  . 
«j  T  a        b        c 

Wlien  there  are  ^;  such  solids  placed  on  each  other  as  above 
described,  the  height  of  the  centre  of  gravity  above  the  plane 
will  be  jjicr ;  and,  if  p  be  the  distance  of  the  foot  of  the  per- 
pendicular drawn  from  the  centre  of  gravity  on  the  horizontal 
plane  from  the  centre  of  the  section,  p  the  same  distance  when 
there  is  but  one  solid,  we  have  p  =  i^p :  the  condition  that  the 
p  solids  shall  not  fall  over  is,  that  pp  shall  be  equal  to  or 
less  than  the  radius  vector  of  the  section  through  the  foot 
of  the  said  perpendicular. 

The  equation  to  the  cutting  plane  is 

Ix  +  my  +  nz  =  vj  (l) ; 

and  if  a,  /8,  7,  be  the  coordinates  of  the  centre  of  the  section, 
a,  /8,  7,  are  subject  to  the  conditions  (see  Gregory's  Solid 
Geometry^  Art.  121) 

la  +  myS  -f  W7  =  ■z^, 

a         /9         7  -ra-  ,  s 


and 


ciH      U'm      c'n      a'l' +  H'm' +  c^i' 


The  coordinates  of  the  foot  of  the  perpendicular  on  (1)  are 
IzTj  v/jCT,  w-ct;  hence  the  equations  to  the  radius  vector  of  the 
section  through  this  foot  are 


y  -  ^    _    g  -  7    _ 


=  p,r  suppose (3), 


Izs  —  o.       wa  —  ^       7*^  —  7 

where  r  is  the  distance  of  the  point  {xyz)  fi'om  (a/37). 

If  we  substitute  these  values  of  ic,  y,  2,  in  the  equation  ta 
the  ellipsoid 

x'      f  _^  g'  _  1 
i?  +  6^  +  c^-*' 


390  SOLUTIONS  OF  SENATE-HOUSE   PROBLEMS. 

we  shall  find  the  length  of  the  radius  vector  r  of  the  section 
through  the  foot  of  the  perpendicular, 

{{1^  -  a)  fir -\- aY       {(yn^  - /3)  fir  + /3Y  _^  {(»^  -  7)  Z^^  +  tF  _  1 
7?  ^  h'  +  6'  '~^' 

the  roots  of  this  equation  are  equal ; 

Now,  from  equations  (2), 
la      1)1/3      7i<y 


a'  "^   F   "^  6'       aT  +  bW  +  6'd' ' 
and    -5  +  7T  +  -ii  =  -^7F 


and  from  equations  (3), 

—  =  (?zj  -  a)'  -I-  [mzj  -  /Sy  +  {nzj  -  7)"' 

=  w"''  -  2^'  +  a'"'  +  ^'  +  7' 
=  d'  +  fi'  +  i'  -  ^' 

Hence  equation  (4)  becomes 

{d'  "^  U'  "^  c"^  ^  dT  +  yW  +  d'd']  p'  ~         dT  +  hW  -f  c^;i^  • 
Now  the  equation  of  equilibrium  is 


p' 

1 

^ 

or 

< 

r 

P' 

henee  the  required  condition  is 


=  or  <  —,  idH''  +  Z^'^y/i''  4  c'n^  -  to''). 


or  |y  (^1  +  ^^  +  ~)  [d^l"  +  h\i^  -f  6^d^)  -  (/  -  i)|  to' 
=  or  <  dP  +  Z/^wi'^  +  c'/i"'. 


APPENDIX.  391 

3.  If  a  plane  area,  bounded  by  a  parabola  and  its  double 
ordinate,  be  supported  by  an  axis  through  the  focus  and  a 
vertical  force  acting  along  the  ordinate,  find  what  portion  may 
be  cut  oif  by  a  line  through  the  focus  without  aifccting  the 
vertical  force ;  and  the  least  area  for  which  this  is  possible. 

Let  P8Q  (fig.  127)  be  the  line  cutting  off  the  portion  PQR : 
the  centre  of  gravity  of  PQR  must  lie  in  the  vertical  through  8^ 
or  if  we  draw  R  V  the  diameter  of  PQ^  and  SG  vertical  meeting 
it  in  O^  G  must  be  the  centre  of  gravity  of  PQR.  Hence 
GV=IRV.     Let  AS=l,  LPSB=e', 


Also 


But 


or 


tan^  = 

21 

SG' 

GV  = 

SG  cot^  = 

2l  coi'6 

. 

SP  = 

21 

1  -  cos^' 

.  PV  = 

SP-  SV  = 

-.  SP- 

SG  cosec 

0 

(      1 

cos^X 

sin'''<9J 

21 

Vl-cos^ 

21 

^xxi'd' 

PV  = 

sni  d 

4Z 
~  sin' 6' 

.5lcofe: 

AP 

Al.5lcoed\ 

! 

.:  cos''^  =  i, 

which  determines  the  position  of  the  line  P8Q. 

If  the  bounding  ordinate  have  its  extremity  C  nearer  the 
vertex  than  the  point  P  just  determined,  let  Q'SP'  be  the  po- 
sition of  the  cutting  line,  the  centre  of  gravity  of  Q'  CP'  must 
lie  in  the  vertical  through  S:  draw  Q'u  vertical:  let  AB'  =  a] 

.'.    I     {2P-{x  +  l)i-xtand}xdx 

=  p^^{2Z4  {I  -x)i  +  x  tan  e]xdx  +  2(^    2  (Ix)^  x  dx. 

J "  Usectf 


392  SOLUTIONS   OF   SENATE-HOUSE   PROBLEMS. 

This  equation,  when  reduced,  will  detennine  the  value  of  6 
in  teniis  of  a. 

The  least  area  for  which  this  is  possible  will  evidently  be 
sucli,  that  the  part  cut  off  will  be  the  half,  and  the  cutting  line 
A  SB:  in  this  case  S  is  the  centre  of  gravity  of  the  whole  area, 
and  AB  =  |Z. 

The  first  part  of  the  Prob.  5,  on  p.  212,  may  be  proved  by 
referring  to  the  values  of  the  angles  contained  between  any 
two  adjacent  sides  of  a  regular  polyhedron  (see  Hall's  Sjjherical 
Trifjonometry^  Art.  59) :  it  appears  that  this  angle  is  a  sub- 
multiple  of  27r  only  in  the  case  of  the  cube. 


THE    END. 


CAMBRIDGE: 
PRINTED    BY    METCALFE    AND    PALMER. 


Plair    1 


MOctHH  Pmimttr  IJtm  Cmmiryjf, 


' 

c. 

f\ 

127. 

f" 

p 

> 

»' 

% 

\^ 

---- 

MttaiA  VrMi-4r.  liHa 


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Cicero  on  Old  Age. 

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